THE FOUR π PUZZLE



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version update of 18 Feb 2004 [176+ kilobytes]

JONATHAN'S MATH PAGES

: annotated hotlinks to other Math Pages What numbers can be made with four copies of the number "pi", or π? Believe it or not: every integer under 1000. And maybe every integer. Read on! Jump immediately to Table of Numbers Represented with 4 copies of π

Introduction

π is the ratio of the circumference of a circle to its diameter. π = 3.141592653589793238462643383279... and has decimal digits which go on forver without repeating. π is one of the most important numbers of all. A good place to start on researching π is: Pi at Mathworld Now, let's have fun with π! "The Weekly Dispatch" of 4 February 1900, which ran a puzzle column by Dudeney, introduced a problem which is still provoking interest today. That problem was the Four Nines Puzzle That was based on the even older Four Fours Puzzle, which is discussed more fully near the bottom of this web page. How can we construct the smaller whole numbers, under 100 for instance, using only all four copies of the number π, parentheses, and the arithmetic operators

"+","-","x","/"?

This old puzzle can be updated if we allow the use of exponentiation, radicals (especially the square root "sqrt"), factorial "!", and the floor function "|_ N _|" and ceiling function "|- N -|". Each of these is explained near the bottom of this web page, but let me give a quick summary, before we see the table of answers and still unsolved numbers.

Factorial: N Factorial is symbolized N!

1! = 1 2! = 1 x 2 = 2 3! = 1 x 2 x 3 = 6 4! = 1 x 2 x 3 x 4 = 24 5! = 1 x 2 x 3 x 4 x 5 = 120 6! = 1 x 2 x 3 x 4 x 5 x 6 = 720 7! = 1 x 2 x 3 x 4 x 5 x 6 x 7 = 5040 8! = 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 = 40,320 9! = 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 = 362,880 10! =1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 x 10 = 3,628,800 ... and so on

Exponentiation: Raising to a Power

Rather than use superscripts, the usual typographical way to show exponents, we use the symbolism that A^B means A to the power of B. For any number N: N^0 = 1 N^1 = N N^2 = N x N N^3 = N x N x N N^4 = N x N x N x N N^5 = N x N x N x N x N ... and so on

Here is a table of π to various powers, rounded down

|_π^0_| = 1 |_π^1_| = |_π_| = 3 |_π^2_| = |_π x π_| = 9 |_π^3_| = |_π x π x π_| = 31 |_π^4_| = |_π x π x π x π_| = 97 |_π^5_| = |_π x π x π x π x π_| = 306 |_π^6_| = |_π x π x π x π x π x π_| = 961 |_π^7_| = |_π x π x π x π x π x π x π_| = 3020 |_π^8_| = |_π x π x π x π x π x π x π x π_| = 9488 |_π^9_| = 29809 |_π^10_| = 93648 |_π^11_| = 294204 |_π^12_| = 924269 |_π^13_| = 2903677 |_π^14_| = 9122171 |_π^15_| = 28658145 |_π^16_| = 90032220 |_π^17_| = 282844563 |_π^18_| = 888582403 |_π^19_| = 2791563949 |_π^20_| = 8769956796 |_π^21_| = 27551631842 |_π^22_| = 86556004191 |_π^23_| = 271923706893 |_π^24_| = 854273519913 |_π^25_| = 2683779414317 |_π^26_| = 8431341691876 |_π^27_| = 26487841119103 |_π^28_| = 83214007069229 |_π^29_| = 261424513284460 |_π^30_| = 821289330402748

Definition of Floor Function and Ceiling Function

Note: the floor function, symbolized as floor(x) = |_ x _| and ceiling function , symbolized as ceiling(x) = |- x -| work on numbers which are not integers, by rounding up or rounding down. That is: |_ 0.5 _| = 0 |- 0.5 -| = 1 |_ 1.2345 _| = 1 |- 1.2345 -| = 2 |_ π _| = |_ 3.141592653689793... _| = 3 |- π -| = |- 3.141592653689793... -| = 4 These last examples suggest our "four π" problem. Note that we don't actually need the ceiling(x) = |- x -| because the same thing can be done with the floor function as follows: ceiling(x) = |- x -| = the least integer greater than or equal to x = - floor(-x) = - |_ -x _|

Jump to a discussion of some deeper mathematical issues related to the Four π Puzzle

Can you improve these below, and/or fill in some blanks? The first blanks still to be solved (as of 18 Feb 2004) are

beyond 1000

... In the table below, whenever I show a number made with fewer than four π, you may use it as raw material to build other numbers, or just multiply by as many copies of 1 as you need to use all four π (because, as it shows near the top, you can make 1 with a single π).

Table of Numbers Represented with 4 copies of π

0 = (π x π) - (π x π) 1 = (π/π) x (π/π) = (π+π) / (π+π) = (πxπ) / (πxπ) = |_ sqrt π _| {which gives us 1 with a single π} 2 = (π/π) + (π/π) = |- sqrt π -| {which gives us 2 with a single π} = |_ 2^(π-2) _| 3 = (π + π + π) / π = |_ π _| {which gives us 3 with a single π} = |_ 3^(π-2) _| = |_ π^(π-2) _| 4 = |- π -| {using just one pi} = |_ 2^(π-1) _| = |_ 4^(π-2) _| 5 = (|_ π _| + |_ π _|) - (π/π) = |_ sqrt |_ sqrt (|_ π _| !)! _| _| {that is, |_ π _|! = 3! = 6, and 6! = 720, and sqrt 720 = 26.83281573..., and sqrt 26.83281573 = 5.18004012822, which rounds down to 5, thus giving us 5 from a single &pi} 6 = 3! = 3x2x1 = |_ π _|! = (|_ π _| + |_ π _|) x (π/π) = |_ 5^(π-2) _| 7 = (|_ π _| + |_ π _|) + (π/π) = |_ π _| + |- π -| {two π} = |_ 6^(π-2) _| = |_ |_π_|^(π-2) _| = |_ sqrt |_ sqrt |_ sqrt |_ sqrt |_ sqrt |_ sqrt 50! _| _| _| _| _| _| {and, as we'll see later, 50 can be made with a single π, hence 7 can be made with a single &pi} 8 = (|_ π _| x |_ π _|) - (π/π) = |- π -| + |- π -| {uses just two π} 9 = (|_ π _| x |_ π _|) x (π/π) = |_ π _| x |_ π _| = |_ π x π _| {with two π} = |_ 7^(π-2) _| = 10 = |- π x π -| {with two π} = (|_ π _| x |_ π _|) + (π/π) = |_ 3^(π-1) _| = |_ |_π_|^(π-1) _| = |_ 8^(π-2) _| = |_ sqrt 5! _| {that is, 10 = the rounded-down square root of 5!=120, and we can make a 5 from a single π; hence we make 10 using only one π} 11 = {10 using only one π} + |_ sqrt π _| = (|_ π _| x |- π -|) - (π/π) = |_ π^(π-1) _| = |_ |_π_|^(π-1) _| 12 = (|_ π _| x |- π -|) x (π/π) = |_ π _| x |- π -| {with two π} = |_ 9^(π-2) _| = |_ |_ π x π _|^(π-2) _| 13 = (|_ π _| x |- π -|) + (π/π) = |_ 10^(π-2) _| = |_ {10 made from one π}^(π-2) _| = {10 made from one π} + |_π_| 14 = |_ π _| + |- π -| + |_ π _| + |- π -| = |- π^(9/4) -| = |- π^((|_ π x π _|)/|_π_|) -| {with three π} 15 = |_ π _| + |- π -| + |- π -| + |- π -| 16 = |- π -| + |- π -| + |- π -| + |- π -| = |- π -| x |- π -| {using just two π} 17 = (|- π -| x |- π -|) + (π/π) = |_ 12^(π-2) _| = |_ (|_ π _| x |- π -|)^(π-2) _| = |_ π^(5/2) _| {using 5 with a single π, 17 uses two π} = |_ 2^(π + 1) _| = |_ |- sqrt π -| ^ (|_π_| + 1) _| = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 100! _| {the rounded-down 128th-root of 100!; since we can make 100 with four &pi this gives us another way to make 17 with four &pi} 18 = |_ |_π_|! x |_π_|!! _| {with only 2 π} {since 6xπ = 18.849555...} = (|- π -| x |- π -|) + |_ sqrt π _| + |_ sqrt π _| = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 102! _| {the rounded-down 128th-root of 102!; since we can make 102 with four &pi, this gives us another way to make 18 with four &pi} 19 = |- |_π_|! x |_π_|!! -| {with only 2 π} {since 6xπ = 18.849555...} = (|- π -| x |- π -|) + |_ sqrt π _| + |- sqrt π -| = |_ 4^(π-1) _| = |_ |-π-|^(π-1) _| = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 103! _| {the rounded-down 128th-root of 103!; since we can make 103 with four nines, this gives us another way to make 19 with four nines} 20 = 4! - 4 = |-π-|! - |-π-| {=20 with just two π} = 16 + 3 + 1 = (|- π -| x |- π -|) + |_π_| + |_ sqrt π_| = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 105! _| {the rounded-down 128th-root of 105!; since we can make 103 with four π, this gives us another way to make 20 with four π} 21 = (|- π -| x |- π -|) + |_π_| + |- sqrt π-| = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 106! _| {the rounded-down 128th-root of 106!; since we can make 106 with four π, this gives us another way to make 21 with four π} 22 = (|- π -| x |- π -|) + |_π_| + |_π_| = |_ 15^(π-2) _| = |_ (|_ π _| x {5 with one π})^(π-2) _| = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 107! _| {the rounded-down 128th-root of 107!; since we can make 107 with four π, this gives us another way to make 22 with four π} 23 = (|- π -| x |- π -|) + |_π_| + |-π-| = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 109! _| {the rounded-down 128th-root of 109!; since we can make 109 with four π, this gives us another way to make 23 with four π} 24 = |- π -| x (|_π_| + |_π_|) {with 3 π) = |-π-|! = 4! {uses just a single π} = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 103! _| {the rounded-down 128th-root of 110!; since we can make 110 with four π, this gives us another way to make 24 with four π} 25 = (|- π -| x (|_π_| + |_π_|)) + |_ sqrt π_| = 4! + 1 = |-π-|! + |_ sqrt π _| = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 111! _| {the rounded-down 128th-root of 111!; since we can make 111 with four π, this gives us another way to make 25 with four π} 26 = (|- π -| x (|_π_| + |_π_|)) + |- sqrt π -| = |_ sqrt |_π_|!)! _| {rounded down square root of (3!)! = 6! =720, to make 26 with a single π} = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 103! _| {the rounded-down 128th-root of 112!; since we can make 112 with four π, this gives us another way to make 26 with four π} 27 = |_π_| x |_π_| x |_π_| = |_π_| ^ |_π_| = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 113! _| {the rounded-down 128th-root of 113!; since we can make 113 with four π, this gives us another way to make 27 with four π} 28 = (|_π_| x |_π_| x |_π_|) + |_ sqrt π_| = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 114! _| {the rounded-down 128th-root of 114!; since we can make 114 with four π, this gives us another way to make 28 with four π} 29 = (|_π_| x |_π_| x |_π_|) + |- sqrt π-| = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 115! _| {the rounded-down 128th-root of 115!; since we can make 115 with four π, this gives us another way to make 29 with four π} 30 = (|_π_| x |_π_| x |_π_|) + |_ π_| = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt {26! x 103!} _| {since we can make each of 26 and 103 with one π, we have 30 with two π} = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 116! _| {the rounded-down 128th-root of 116!; since we can make 116 with four π, this gives us another way to make 30 with four π} 31 = |_ π x π x π _| {with three π} = |_ π ^ |_ π _| _| {with two π} = |_ |_π_| ^ π _| {with two π} = |_ 5^(π-1) _| = |_ {5 with one π}^(π-1) _| = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 117! _| {the rounded-down 128th-root of 117!; since we can make 117 with four π, this gives us another way to make 31 with four π} 32 = |- π x π x π -| {with three π} = |- π ^ |_ π _| -| {with two π} = |_ |_π_| ^ π _| {with two π} 33 = |- π x π x π -| + |_ sqrt π _| = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 118! _| {the rounded-down 128th-root of 118!; since we can make 118 with four π, this gives us another way to make 33 with four π} 34 = |- π x π x π -| + |- sqrt π -| = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 119! _| {the rounded-down 128th-root of 119!; since we can make 119 with four π, this gives us another way to make 34 with four π} 35 = |- π x π x π -| + |_ π _| = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 120! _| {the rounded-down 128th-root of 103!; since we can make 120 with a single π, (|_ π _|!)! this gives us another way to make 35 with one π} = |_ 2^(2+π) _| = |_ |- sqrt π -|^(|- sqrt π -| + π) _| 36 = |_ π ^ π _| = |_π_|! x |_π_|! {with just two π} 37 = |- π ^ π -| {with two π} = |_ 24^(π-2) _| = |_ (|-π-|!)^(π-2) _| = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 121! _| {the rounded-down 128th-root of 121!; since we can make 121 with four π, this gives us another way to make 19 with four π} 38 = |_ π ^ π _| + |_ sqrt π _| + |_ sqrt π _| = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 122! _| {the rounded-down 128th-root of 122!; since we can make 122 with four π, this gives us another way to make 38 with four π} 39 = |_ π ^ π _| + |_ π_| = |_ sqrt sqrt sqrt sqrt sqrt 71! _| {and, as we see above, 71 can be made with a single π, hence 39 can be made with a single π} 40 = |_ π ^ π _| + |_ π_| + |_ sqrt π _| = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 123! _| {the rounded-down 128th-root of 123!; since we can make 123 with four π, this gives us another way to make 40 with four π} 41 = |_ π ^ π _| + |_ π_| + |- sqrt π -| = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 124! _| {the rounded-down 128th-root of 124!; since we can make 124 with four π, this gives us another way to make 41 with four π} 42 = |_ π ^ π _| + |_ π_| + |_ π_| 43 = |_ sqrt |_ sqrt 10! _| _| {and, as we see above, 10 can be made with a single π, hence 43 can be made with a single π} {that is, the rounded-down square root of 10! = sqrt 3628800 = 1904.940943... which rounds down to 1904, and sqrt 1904 = 43.6348484585..., which rounds down to 43} = |_ sqrt |_ sqrt |_ sqrt |_ sqrt |_ sqrt ((|_π_|!)!)! _| _| _| _| = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 125! _| {the rounded-down 128th-root of 125!; since we can make 125 with four nines, this gives us another way to make 43 with four nines} 44 = |_ sqrt |_ sqrt |_ sqrt |_ sqrt |_ sqrt 43! _| _| _| _| _| {and, as we see above, 43 can be made with a single π, hence 44 can be made with a single π} = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 126! _| {the rounded-down 128th-root of 126!; since we can make 126 with four π, this gives us another way to make 44 with four π} 45 = |_ π ^ π _| + |_ π x π _| = |_ π^(10/3) _| = |_ π^({10 with one π}/|_π_|) _| {with two π} 46 = |_ π ^ π _| + |- π x π -| = |_ 6^(π-1) _| = |_ |_π_|!^(π-1) _| = |_ sqrt (|_π_|!! x |_π_|) _| {with two π} {because 6!x3 = 720x3 = 2160, and sqrt 2160 = 46.47...} = |_ sqrt |_ sqrt |_ sqrt |_ sqrt 26! _| _| _| _| {and, as we see above, 26 can be made with a single π, hence 46 can be made with a single π} = |_ sqrt sqrt sqrt sqrt 26! _| {since we can make 26 with one π, this gives 2116 with one π} = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 127! _| {the rounded-down 128th-root of 127!; since we can make 127 with four π, this gives us another way to make 46 with four π} 47 = ((|- π-| !) x |- sqrt π -|) - π/π = |- sqrt (|_π_|!! x |_π_|) -| {with two π} {because 6!x3 = 720x3 = 2160, and sqrt 2160 = 46.47...} 48 = (|- π-| !) x |- sqrt π -| {with two π} = |_ 30^(π-2) _| = |_ (|_π_| x {5 with one π})^(π-2) _| = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 128! _| {the rounded-down 128th-root of 128!; since we can make 128 with four π, this gives us another way to make 48 with four π} 49 = (|_ π_| + |- π-|) x (|_ π_| + |- π-|) = ((|- π-| !) x |- sqrt π -|) + π/π = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt {43! x 103!} _| {since we can make each of 43 and 103 with one π, we have 49 with two π} 50 = 24 + 24 + 2 = 4! + 4! + 2 = |-π-|! + |-π-|! + |- sqrt π -| = ((|- π-| !) x |- sqrt π -|) + |_π_| + |_π_| = |_ 31^(π-2) _| = |_ (|_ |_π_| ^ π _|)^(π-2) _| = |_ sqrt |_ sqrt |_ sqrt |_ sqrt 44! _| _| _| _| {and, as we see above, 44 can be made with a single π, hence 50 can be made with a single π} = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 129! _| {the rounded-down 128th-root of 129!; since we can make 129 with four π, this gives us another way to make 50 with four π} 51 = ((|- π-| !) x |- sqrt π -|) + |_π_| + |-π-| = 24 + 24 + 3 = 4! + 4! + 2 = |-π-|! + |-π-|! + |_π_| = |- sqrt sqrt sqrt sqrt sqrt 44! -| {and, as we see above, 44 can be made with a single π, hence 51 can be made with a single π} 52 = |_ sqrt |_π_|!)! _| + |_ sqrt |_π_|!)! _| {with two π} = |_ 32^(π-2) _| = |_ (|- |_π_| ^ π -|)^(π-2) _| = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 130! _| {the rounded-down 128th-root of 130!; since we can make 130 with four π, this gives us another way to make 52 with four π} 53 = |_ sqrt |_π_|!)! _| + |_ sqrt |_π_|!)! _| + |_sqrt π_| {with three π} = |_ sqrt sqrt sqrt sqrt sqrt sqrt {43! x 46!} _| {since we can make each of 43 and 46 with one π, we have 53 with two π} = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt {46! x 103!} _| {since we can make each of 46 and 103 with one π, we have 53 with two π} 54 = |_ π_| x |_ π_| x |_ π_| x |-sqrt π-| = |_ π^(7/2) _| = |_ π^((|_π_| + |-π-|)/|-sqrt π-|) _| {three π} = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 131! _| {the rounded-down 128th-root of 131!; since we can make 131 with four π, this gives us another way to make 54 with four π} 55 = |- π^(7/2) -| = |- π^((|_π_| + |-π-|)/|-sqrt π-|) -| {three π} = 5 x 11 = |_ sqrt |_ sqrt (|_ π _| !)! _| _| x ((|_ π _| x |- π -|) - |_sqrt π_|) 56 = (|_ π _| + |- π -|) x (|- π -| + |- π -|) = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 132! _| {the rounded-down 128th-root of 132!; since we can make 132 with four π, this gives us another way to make 56 with four π} 57 = 3 x 19 = |_π_| x ((|-π-| x |-π-|) + |_π_|) 58 = 60 - 2 = (|_π_|! x |_π_|!) + |-π-|! - |- sqrt π -| = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 133! _| {the rounded-down 128th-root of 133!; since we can make 133 with four π, this gives us another way to make 58 with four π} 59 = 60 - 1 = (|_π_|! x |_π_|!) + |-π-|! - |_sqrt π_| 60 = 36 + 24 = (3! x 3!) + 4! = (|_π_|! x |_π_|!) + |-π-|! {three π} = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt {50! x 103!} _| {since we can make each of 50 and 103 with one π, we have 60 with two π} = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 134! _| {the rounded-down 128th-root of 134!; since we can make 134 with four π, this gives us another way to make 60 with four π} 61 = 60 + 1 = (|_π_|! x |_π_|!) + |-π-|! + |_sqrt π_| 62 = 31 + 31 = |_ π ^ |_ π _| _| + |_ π ^ |_ π _| _| = 60 + 2 = (|_π_|! x |_π_|!) + |-π-|! + |- sqrt π -| 63 = 60 + 3 = (|_π_|! x |_π_|!) + |-π-|! + |_π_| = |_ sqrt |_ sqrt |_ sqrt |_ sqrt |_ sqrt 46! _| _| _| _| _| {and, as we see above, 46 can be made with a single π, hence 63 can be made with a single π} = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 135! _| {the rounded-down 128th-root of 135!; since we can make 135 with four π, this gives us another way to make 63 with four π} 64 = |_ 7^(π-1) _| = 32 + 32 = |- π ^ |_ π _| -| + |- π ^ |_ π _| -| = |- sqrt sqrt sqrt sqrt sqrt 46! -| {since we can make 46 with one π} this gives us 64 with one π} 65 = 60 + 5 = = (|_π_|! x |_π_|!) + |-π-|! + |_ sqrt |_ sqrt (|_ π _| !)! _| _| = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 136! _| {the rounded-down 128th-root of 136!; since we can make 136 with four nines, this gives us another way to make 65 with four nines} 66 = 98 - 32 = |- π ^ 4-| - |- π ^ 3-| = |- π ^ (|-π-|) -| - |- π ^ (|_pi;_|) -| 67 = rounded up square root of 9!/9 = |- (sqrt((|_π_|x|_π_|)!))/(|_π_|x|_π_|) -| = |_ sqrt sqrt sqrt sqrt sqrt sqrt {43! x 50!} _| {since we can make each of 43 and 50 with one π, we have 67 with two π} 68 = {70 made with a single π} - 2 = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 137! _| {the rounded-down 128th-root of 137!; since we can make 137 with four π, this gives us another way to make 68 with four π} 69 = {70 made with a single π} - 1 70 = |_ sqrt 7! _| {and, as we see above, 7 can be made with a single π, hence 70 can be made with a single π} = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 138! _| {the rounded-down 128th-root of 138!; since we can make 138 with four π, this gives us another way to make 70 with four π} 71 = |- sqrt 7! -| {that is, the ceiling function of sqrt 5040} {and, as we see above, 7 can be made with a single π, hence 71 can be made with a single π} 72 = 9 x 8 = (|_ π _| x |_ π _|) x (|- π -| + |- π -|) 73 = 81 - 8 = (3 ^ 4) - 8 = (|_ π _| ^ |- π -|) - |-(π-| - |-(π-| = |_ π^(15/4) _| = |_ π^((|_ π_| x {5 with one π} _|)/|_π_|) _| {with three π} = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 139! _| {the rounded-down 128th-root of 139!; since we can make 139 with four π, this gives us another way to make 73 with four π} 74 = 81 - 7 = (3 ^ 4) - 7 = (|_ π _| ^ |- π -|) - |_(π_| - |-(π-| = |- π^(15/4) -| = |- π^((|_ π_| x {5 with one π} _|)/|_π_|) -| {with three π} 75 = 81 - 6 = (3 ^ 4) - 6 = (|_ π _| ^ |- π -|) - |_(π+π)_| 76 = 81 - 5 = (3 ^ 4) + 5 = (|_ π _| ^ |- π -|) - |-π-| - |_sqrt π_| = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 140! _| {the rounded-down 128th-root of 140!; since we can make 140 with four π, this gives us another way to make 76 with four π} 77 = |_ 4 ^ π _| = |_ |-π-| ^ π _| {with two π} = |_ sqrt sqrt sqrt sqrt sqrt sqrt (9x9)! _| {that is, 77 is the 64th root of 81!, rounded down} = |_ sqrt sqrt sqrt sqrt sqrt sqrt ((|_ π _| x |_ π _|)x(|_ π _| x |_ π _|))! _| 78 = |- 4 ^ π -| = |- |-π-| ^ π -| {with two π} = 81 - 3 = (3 ^ 4) - 3 = (|_ π _| ^ |- π -|) - |_π_| 79 = 81 - 2 = (3 ^ 4) - 2 = (|_ π _| ^ |- π -|) - |- sqrt π -| = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 141! _| {the rounded-down 128th-root of 141!; since we can make 141 with four π, this gives us another way to make 79 with four π} 80 = 81 - 1 = (3 ^ 4) - 1 = (|_ π _| ^ |- π -|) - |_ sqrt π _| 81 = 9 x 9 = (|_ π _| x |_ π _|) x (|_ π _| x |_ π _|) = 3 ^ 4 = |_ π _| ^ |- π -| {with just two π} = |_ sqrt sqrt sqrt sqrt sqrt sqrt {46! x 50!} _| {since we can make each of 46 and 50 with one π, we have 81 with two π} 82 = 81 + 1 = (3 ^ 4) - 1 = (|_ π _| ^ |- π -|) + |_ sqrt π _| = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 142! _| {the rounded-down 128th-root of 142!; since we can make 142 with four π, this gives us another way to make 82 with four π} 83 = 81 + 2 = (3 ^ 4) + 2 = (|_ π _| ^ |- π -|) + |- sqrt π -| 84 = 81 + 3 = (3 ^ 4) + 3 = (|_ π _| ^ |- π -|) + |_π_| 85 = |_ 8^(π-1) _| = |_ (|-π-| + |-π-|)^(π-1) _| = 81 + 4 = (3 ^ 4) + 4 = (|_ π _| ^ |- π -|) + |-π-| = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 143! _| {the rounded-down 128th-root of 143!; since we can make 143 with four π, this gives us another way to make 85 with four π} 86 = 81 + 5 = (3 ^ 4) + 5 = (|_ π _| ^ |- π -|) + |-π-| + |_sqrt π_| 87 = 81 + 6 = (3 ^ 4) + 6 = (|_ π _| ^ |- π -|) + |_π_| + |_π_| 88 = 81 + 7 = (3 ^ 4) + 7 = (|_ π _| ^ |- π -|) + |_π_| + |-π-| 89 = 81 + 8 = (3 ^ 4) + 8 = (|_ π _| ^ |- π -|) + |-π-| + |-π-| = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 144! _| {the rounded-down 128th-root of 144!; since we can make 144 with two π, this gives us another way to make 89 with two π} 90 = 81 + 9 = (3 ^ 4) + 9 = (|_ π _| ^ |- π -|) + (|_π_| x |_π_|) 91 = 81 + 10 = (3 ^ 4) + 10 = (|_ π _| ^ |- π -|) + {10 with a single π} = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt {63! x 103!} _| {since we can make each of 63 and 103 with one π, we have 91 with two π} 92 = 46 x 2 = |_ sqrt (|_π_|!! x |_π_|) _| x |- sqrt π -| {with three π} {because 6!x3 = 720x3 = 2160, and sqrt 2160 = 46.47...} = 81 + 11 = (3 ^ 4) + 11 = (|_ π _| ^ |- π -|) + {10 with a single π} + |_ sqrt π _| = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 145! _| {the rounded-down 128th-root of 145!; since we can make 145 with four nines, this gives us another way to make 92 with four nines} 93 = 81 + 12 = (3 ^ 4) + 12 = (|_ π _| ^ |- π -|) + (|_ π _| x |- π -|) = 31 x 3 = |_ π ^ |_ π_| _| x |_ π_| {with three π} 94 = |_ 3^(π + 1) _| = |_ |_π_| ^ (|_π_| + 1) _| = 99 - 5 = |_ π ^ |-π-| _| + |- sqrt π -| - |_ sqrt |_ sqrt (|_ π _| !)! _| _| 95 = 99 - 4 = |_ π ^ |-π-| _| + |- sqrt π -| - |- π-| 96 = 99 - 3 = |_ π ^ |-π-| _| + |- sqrt π -| - |_ π_| = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 146! _| {the rounded-down 128th-root of 146!; since we can make 146 with four π, this gives us another way to make 96 with four π} 97 = |_ π x π x π x π _| = |_ π ^ |-π-| _| {with two π} 98 = |- π x π x π x π -| = |- π ^ (|-π-|) -| {with two π} {that is, 98 is the rounded up π to the 4th power} 99 = |_ π ^ |-π-| _| + |- sqrt π -| {with three π} 100 = |- π x π -| x |- π x π -| = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 147! _| {the rounded-down 128th-root of 147!; since we can make 147 with four nines, this gives us another way to make 100 with four nines} 101 = 99 + 2 = |_ π ^ |-π-| _| + |- sqrt π -| + |- sqrt π -| 102 = 99 + 3 = |_ π ^ |-π-| _| + |- sqrt π -| + |_ π_| 103 = 99 + 4 = |_ π ^ |-π-| _| + |- sqrt π -| + |- π-| = |_ sqrt |_ sqrt |_ sqrt |_ sqrt |_ sqrt 50! _| _| _| _| _| {and, as we see above, 50 can be made with a single π, hence 103 can be made with a single π} = |_ sqrt sqrt sqrt sqrt sqrt 50! _| {and, as we see above, 50 can be made with a single π, hence 103 can be made with a single π} {verifed elimination of 4 floors} 104 = 99 + 5 = |_ π ^ |-π-| _| + |- sqrt π -| + |_ sqrt |_ sqrt (|_ π _| !)! _| _| = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 148! _| {the rounded-down 128th-root of 148!; since we can make 148 with four π, this gives us another way to make 104 with four π} 105 = 99 + 3! = |_ π ^ |-π-| _| + |- sqrt π -| + |_ π_|! 106 = 97 + 9 = |_ π ^ |-π-| _| + (|_π_| x |_π_|) 107 = (36 x 3) - 1 = (|_ π ^ π _| x |_π_|) - |_ sqrt π _| = 97 + 10 = |_ π ^ |-π-| _| + {10 with one π} {uses 3 π} 108 = 36 x 3 = |_ π ^ π _| x |_π_| {with three π} = 97 + 10 + 1 = |_ π ^ |-π-| _| + {10 with one π} + |_ sqrt π_| = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 149! _| {the rounded-down 128th-root of 149!; since we can make 149 with four π, this gives us another way to make 108 with four π} 109 = (36 x 3) + 1 = (|_ π ^ π _| x |_π_|) + |_ sqrt π _| = {99 with three π} + {10 with one π} 110 = |_ 9^(π-1) _| = |_ (|_ π x π _|)^(π-1) _| = 97 + 10 + 3 = |_ π ^ |-π-| _| + {10 with one π} + |_ π_| = (36 x 3) + 2 = (|_ π ^ π _| x |_π_|) + |- sqrt π -| 111 = 5! - 9 = (|_ sqrt |_ sqrt (|_ π _| !)! _| _|)! - (|_π_| x |_π_|) = 97 + 10 + 4 = |_ π ^ |-π-| _| + {10 with one π} + |- π-| = (36 x 3) + 3 = (|_ π ^ π _| x |_π_|) + |_π_| 112 = 5! - 8 = (|_ sqrt |_ sqrt (|_ π _| !)! _| _|)! - |-π-| - |-π-| = (36 x 3) + 4 = (|_ π ^ π _| x |_π_|) + |-π-| = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 103! _| {the rounded-down 128th-root of 150!; since we can make 150 with four nines, this gives us another way to make 112 with four nines} 113 = 5! - 7 = (|_ sqrt |_ sqrt (|_ π _| !)! _| _|)! - |_π_| - |-π-| 114 = 5! - 6 = (|_ sqrt |_ sqrt (|_ π _| !)! _| _|)! - |_π_| - |_π_| = |_ π^(π + 1) _| = |_ |_π_| ^ (|_π_| + 1) _| 115 = 5! - 5 = (|_ sqrt |_ sqrt (|_ π _| !)! _| _|)! - (|_ sqrt |_ sqrt (|_ π _| !)! _| _|) = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt {70! x 103!} _| {since we can make each of 70 and 103 with one π, we have 115 with two π} 116 = 5! - 4 = (|_ sqrt |_ sqrt (|_ π _| !)! _| _|)! - |-π-| 117 = 5! - 3 = (|_ sqrt |_ sqrt (|_ π _| !)! _| _|)! - |_π_| = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 151! _| {the rounded-down 128th-root of 151!; since we can make 151 with four nines, this gives us another way to make 117 with four nines} 118 = 5! - 2 = (|_ sqrt |_ sqrt (|_ π _| !)! _| _|)! - |- sqrt π -| 119 = 5! - 1 = (|_ sqrt |_ sqrt (|_ π _| !)! _| _|)! - |_ sqrt π _| = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt {70! x 103!} _| {since we can make each of 71 and 103 with one π, we have 119 with two π} 120 = 5! = (|_ sqrt |_ sqrt (|_ π _| !)! _| _|)! {with a single π} 121 = 5! + 1 = (|_ sqrt |_ sqrt (|_ π _| !)! _| _|)! + |_ sqrt π _| 122 = 5! + 2 = (|_ sqrt |_ sqrt (|_ π _| !)! _| _|)! + |- sqrt π -| = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 152! _| {the rounded-down 128th-root of 152!; since we can make 152 with four π, this gives us another way to make 122 with four π} 123 = 5! + 3 = (|_ sqrt |_ sqrt (|_ π _| !)! _| _|)! + |_π_| 124 = 5! + 4 = (|_ sqrt |_ sqrt (|_ π _| !)! _| _|)! + |-π-| = 31 x 4 = |_ π ^ |_ π_| _| x |- π-| {with three π} 125 = 5 ^ 3 = (|_ sqrt |_ sqrt (|_ π _| !)! _| _|) ^ |_π_| {with three π} 126 = 5! + 6 = (|_ sqrt |_ sqrt (|_ π _| !)! _| _|)! + |_π_| + |_π_| 127 = ( |_π_|! - |_ sqrt |_π_| _| )! + |_π_|! + |_ sqrt π _| = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 153! _| {the rounded-down 128th-root of 153!; since we can make 153 with four π, this gives us another way to make 127 with four π} 128 = 5! + 8 = (|_ sqrt |_ sqrt (|_ π _| !)! _| _|)! + |-π-| + |-π-| 129 = 5! + 9 = (|_ sqrt |_ sqrt (|_ π _| !)! _| _|)! + (|_π_| x |_π_|) 130 = 5! + 10 = (|_ sqrt |_ sqrt (|_ π _| !)! _| _|)! + {10 with one π} 131 = 5! + 11 {to be done} = |_ sqrt sqrt sqrt sqrt 31! _| {since we can make 31 with two π} this gives us 131 with two π} 132 = 5! + 12 = (|_ sqrt |_ sqrt (|_ π _| !)! _| _|)! + (|_ π _| x |- π -|) = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 154! _| {the rounded-down 128th-root of 154!; since we can make 154 with four π, this gives us another way to make 132 with four π} 133 = |- 3rd-root-of-9!) -| that is, ceiling of cube root of 9! = (9!)^(1/3) = ((|_π_| x |_π_|)!) ^ ((|_ sqrt π _|) / |_π_|) = 97 + 36 = |_ π ^ (|-π-|) _| + (|_π_|! x |_π_|!) 134 = 98 + 36 = |- π ^ (|-π-|) -| + (|_π_|! x |_π_|!) = |_ sqrt(sqrt((9-to-the-9))) _| - ((sqrt 9) + (sqrt 9)) 135 = 5! + 4! - 9 {to be done} = |- sqrt(sqrt((9-to-the-9))) -| - ((sqrt 9) + (sqrt 9)) [ceiling] 136 = 5! + 4! - 8 {to be done} 137 = 5! + 4! - 7 {to be done} = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 155! _| {the rounded-down 128th-root of 155!; since we can make 155 with four π, this gives us another way to make 137 with four π} 138 = |_ 10^(π-1) _| = |_ {10 with one π}^(π-1) _| = 5! + 4! - 6 = (|_ sqrt |_ sqrt (|_ π _| !)! _| _|)! + |- π -|! - |_π_|! 139 = 5! + 4! - 5 {to be done} = |_ 3 x sqrt (6!x3) _| = |_ |_π_| x |_ sqrt (|_π_|!! x |_π_|) _| {with three π} {because 6!x3 = 720x3 = 2160, and sqrt 2160 = 46.47...} 140 = |_ sqrt(sqrt((9^9))) _| = 5! + 4! - 4 = (|_ sqrt |_ sqrt (|_ π _| !)! _| _|)! + |- π -|! - |-π-| 141 = 5! + 4! - 3 = (|_ sqrt |_ sqrt (|_ π _| !)! _| _|)! + |- π -|! - |_π_| 142 = |- sqrt(sqrt((9^9))) -| = |- sqrt(sqrt(((|_ π x π _| ^ |_ π x π _|))) -| = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 156! _| {the rounded-down 128th-root of 156!; since we can make 156 with four π, this gives us another way to make 142 with four π} = 5! + 4! - 2 = (|_ sqrt |_ sqrt (|_ π _| !)! _| _|)! + |- π -|! - |- sqrt π -| 143 = 5! + 4! - 1 = (|_ sqrt |_ sqrt (|_ π _| !)! _| _|)! + |- π -|! - |_ sqrt π _| 144 = 120 + 24 = 5! + 4! = (|_ sqrt |_ sqrt (|_ π _| !)! _| _|)! + |- π -|! {note that this uses just two π} = 36 x 4 = (|_ π ^ π _| x |-π-|) {with three π} 145 = 5! + 4! + 1 = (|_ sqrt |_ sqrt (|_ π _| !)! _| _|)! + |- π -|! + |_ sqrt π _| 146 = {153 with three π} - 7 = {to be done} = 5! + 4! + 2 = (|_ sqrt |_ sqrt (|_ π _| !)! _| _|)! + |- π -|! + |- sqrt π -| 147 = {153 with three π} - 6 = {153 with three π} - |_π_|! = 5! + 4! + 3 = (|_ sqrt |_ sqrt (|_ π _| !)! _| _|)! + |- π -|! + |_π_| 148 = {153 with three π} - 5 = {153 with three π} - |_ sqrt |_ sqrt (|_ π _| !)! _| _| _| = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 157! _| {the rounded-down 128th-root of 157!; since we can make 157 with four π, this gives us another way to make 148 with four π} = 5! + 4! + 4 = (|_ sqrt |_ sqrt (|_ π _| !)! _| _|)! + |- π -|! + |-π-| 149 = {153 with three π} - |-π-| = 5! + 4! + 5 = (|_ sqrt |_ sqrt (|_ π _| !)! _| _|)! + |- π -|! + (|_ sqrt |_ sqrt (|_ π _| !) 150 = {153 with three π} - |_π_| = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt 9999! _| {that is, 150 is the 16384th root of 9999!, rounded down} = 5! + 4! + 3! = (|_ sqrt |_ sqrt (|_ π _| !)! _| _|)! + |- π -|! + |_ π _|! 151 = {153 with three π} - 2 = {153 with three π} - |- sqrt π -| 152 = {153 with three π} - 1 = {153 with three π} - |_ sqrt π _| 153 = |_ π ^ 5 _| / 2 = |_ π ^ |_ sqrt |_ sqrt (|_ π _| !)! _| _| _| / |- sqrt π -| {using three π} 154 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {43! x 63!} _| {since we can make each of 43 and 63 with one π, we have 154 with two π} = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 158! _| {the rounded-down 128th-root of 158!; since we can make 158 with four π, this gives us another way to make 154 with four π} 154 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {91!} _| {that is, 154 = the 64th root of 91!, so insert an existing four π solution for 91 here} 155 = 31 x 5 = |_ π ^ |_ π _| _| x |_ sqrt |_ sqrt (|_ π _| !)! _| _| {uses three π} 156 = |_ 5 ^ π _| {given 5 with one π, we have 156 with two π} = (31 x 5) + 1 = (|_ π ^ |_ π _| _| x |_ sqrt |_ sqrt (|_ π _| !)! _| _|) + |_ sqrt π _| 157 = |- 5 ^ π -| {given 5 with one π, we have 157 with two π} = (31 x 5) + 2 = (|_ π ^ |_ π _| _| x |_ sqrt |_ sqrt (|_ π _| !)! _| _|) + |- sqrt π -| 158 = (31 x 5) + 3 = (|_ π ^ |_ π _| _| x |_ sqrt |_ sqrt (|_ π _| !)! _| _|) + |_π_| 159 = (31 x 5) + 4 = (|_ π ^ |_ π _| _| x |_ sqrt |_ sqrt (|_ π _| !)! _| _|) + |-π-| 160 = 32 x 5 = |- π ^ |_ π _| -| x |_ sqrt |_ sqrt (|_ π _| !)! _| _|) {with three π} = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 159! _| {the rounded-down 128th-root of 159!; since we can make 159 with four π, this gives us another way to make 160 with four π} 161 = (32 x 5) + 1 = |- π ^ |_ π _| -| x |_ sqrt |_ sqrt (|_ π _| !)! _| _|) + |_ sqrt π _| 162 = (32 x 5) + 2 = (|- π ^ |_ π _| -| x |_ sqrt |_ sqrt (|_ π _| !)! _| _|) + |- sqrt π -| 163 = (32 x 5) + 3 = (|- π ^ |_ π _| -| x |_ sqrt |_ sqrt (|_ π _| !)! _| _|) + |_π_| = |_ sqrt sqrt sqrt sqrt 32! _| {since we can make 32 with two π} this gives us 163 with two π} 164 = (32 x 5) + 4 = (|- π ^ |_ π _| -| x |_ sqrt |_ sqrt (|_ π _| !)! _| _|) + |-π-| 165 = (32 x 5) + 5 = (|- π ^ |_ π _| -| x |_ sqrt |_ sqrt (|_ π _| !)! _| _|) + |_ sqrt |_ sqrt (|_ π _| !)! _| _| 166 = (32 x 5) + 6 = (|- π ^ |_ π _| -| x |_ sqrt |_ sqrt (|_ π _| !)! _| _|) + |_π_|! = |_ sqrt sqrt sqrt sqrt sqrt sqrt {92!} _| {that is, 166 = the 64th root of 92!, so insert an existing four π solution for 92 here} 167 = ??? = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 160! _| {the rounded-down 128th-root of 160!; since we can make 160 with four π, this gives us another way to make 167 with four π} 168 = 84 x 2 = (81 + 3) x 2 = ((3 ^ 4) + 3) x 2 = ((|_ π _| ^ |- π -|) + |_π_|) x |- sqrt π -| 169 = 13 x 13 = ({10 made from one π} + |_π_|) x ({10 made from one π} + |_π_|) 170 = 85 x 2 = (81 + 4) x 2 = ((3 ^ 4) + 4) x 2 = ((|_ π _| ^ |- π -|) + |-π-) x |- sqrt π -| 171 = 3 x 3 x 19 = |_π_| x |_π_| x |- |_π_|! x |_π_|!! -| 172 = |_ π^(9/2) _| = |_ π^((|_ π x π _|)/|-sqrt π-|) _| {with three π} = |_ sqrt sqrt sqrt sqrt sqrt sqrt (99 - 3!)!) _| - 3! 173 = |- π^(9/2) -| = |- π^((|_ π x π _|)/|-sqrt π-|) -| {with three π} = |_ sqrt sqrt sqrt (99^9) _| - |-π-| = |_ ( |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt (9-to-the-(9-to-the-(sqrt 9))) _| ) / sqrt 9 _| 174 = (36 x 5) - 6 = (|_ π ^ π _| x |_ sqrt |_ sqrt (|_ π _| !)! _| _|) - |_π_|! = |_ sqrt sqrt sqrt (99-to-the-9th) _| - |_ sqrt 3 _| = |- |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt (9-to-the-(9-to-the-3)) _| / 3 -| [floor in ceiling] = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 161! _| {the rounded-down 128th-root of 161!; since we can make 161 with four π, this gives us another way to make 174 with four π} 175 = (36 x 5) - 5 = (|_ π ^ π _| x |_ sqrt |_ sqrt (|_ π _| !)! _| _|) - |_ sqrt |_ sqrt (|_ π _| !)! _| _| = |_ sqrt sqrt sqrt sqrt sqrt sqrt (99 - 3!)!) _| - 3 = |_ sqrt sqrt sqrt (99-to-the-9th) _| 176 = (36 x 5) - 4 = (|_ π ^ π _| x |_ sqrt |_ sqrt (|_ π _| !)! _| _|) - |-π-| 177 = (36 x 5) - 3 = (|_ π ^ π _| x |_ sqrt |_ sqrt (|_ π _| !)! _| _|) - |_π_| = |_ sqrt sqrt sqrt sqrt sqrt sqrt (99 - 3!)!) _| - |_ sqrt 3 _| 178 = (36 x 5) - 2 = (|_ π ^ π _| x |_ sqrt |_ sqrt (|_ π _| !)! _| _|) - |- sqrt π -| = |_ sqrt sqrt sqrt sqrt sqrt sqrt {93!} _| {that is, 178 = the 64th root of 93!, so insert an existing four nines solution for 93 here} {note that 93 = 99 - (sqrt 9)!} 179 = (36 x 5) - 1 = (|_ π ^ π _| x |_ sqrt |_ sqrt (|_ π _| !)! _| _|) - |_ sqrt π _| 180 = 36 x 5 = (|_ π ^ π _| x |_ sqrt |_ sqrt (|_ π _| !)! _| _|) {with three π} 181 = (36 x 5) + 1 = (|_ π ^ π _| x |_ sqrt |_ sqrt (|_ π _| !)! _| _|) + |_ sqrt π _| = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 162! _| {the rounded-down 128th-root of 162!; since we can make 162 with four π, this gives us another way to make 181 with four π} 182 = (36 x 5) + 2 = (|_ π ^ π _| x |_ sqrt |_ sqrt (|_ π _| !)! _| _|) + |- sqrt π -| 183 = (36 x 5) + 3 = (|_ π ^ π _| x |_ sqrt |_ sqrt (|_ π _| !)! _| _|) + |_π_| 184 = (36 x 5) + 4 = (|_ π ^ π _| x |_ sqrt |_ sqrt (|_ π _| !)! _| _|) + |-π-| = |_ sqrt sqrt sqrt sqrt sqrt sqrt {46! x 63!} _| {since we can make each of 46 and 63 with one π, we have 184 with two π} 185 = 37 x 5 = (|- π ^ π -| x |_ sqrt |_ sqrt (|_ π _| !)! _| _|) {with three π} = |_ 4 x sqrt 2160 _| = = |_ sqrt (|- sqrt π -| x (|_π_|!!) x |_π_|) )_| {with three π} {because 6!x3 = 720x3 = 2160, and sqrt 2160 = 46.47...} 186 = (37 x 5) + 1 = (|- π ^ π -| x |_ sqrt |_ sqrt (|_ π _| !)! _| _|) + |_ sqrt π _| = |- 4 x sqrt 2160 -| = = |- sqrt (|- sqrt π -| x (|_π_|!!) x |_π_|) )-| {with three π} {because 6!x3 = 720x3 = 2160, and sqrt 2160 = 46.47...} 187 = (37 x 5) + 2 = (|- π ^ π -| x |_ sqrt |_ sqrt (|_ π _| !)! _| _|) + |- sqrt π -| 188 = (37 x 5) + 3 = (|- π ^ π -| x |_ sqrt |_ sqrt (|_ π _| !)! _| _|) + |_π_| = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 163! _| {the rounded-down 128th-root of 163!; since we can make 163 with four π, this gives us another way to make 188 with four π} 189 = (37 x 5) + 4 = (|- π ^ π -| x |_ sqrt |_ sqrt (|_ π _| !)! _| _|) + |-π-| 190 = 200 - 10 = |_ sqrt 8! _| - 10 = |_ sqrt (|- π -| + |- π -|)! _| - {10 with one π} = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt (9-to-the-(3!-to-the-3)) _| - 3! 191 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {94!} _| {that is, 191 = the 64th root of 94!, so insert an existing four π solution for 94 here} 192 = 200 - 8 = |_ sqrt 8! _| - 8 = |_ sqrt (|- π -| + |- π -|)! _| - |-π-| - |-π-| = ( |_ (sqrt(9!))/9 _| x 3 ) - 3! 193 = 200 - 7 = |_ sqrt 8! _| - 7 = |_ sqrt (|- π -| + |- π -|)! _| - |_π_| - |-π-| = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt (9-to-the-(3!-to-the-3)) _| - 3 194 = 200 - 6 = |_ sqrt 8! _| - 6 = |_ sqrt (|- π -| + |- π -|)! _| - |_π_|! 195 = 200 - 5 = |_ sqrt 8! _| - 5 = |_ sqrt (|- π -| + |- π -|)! _| - |_ sqrt |_ sqrt (|_ π _| !)! _| _| 196 = 200 - 4 = |_ sqrt 8! _| - 4 = |_ sqrt (|- π -| + |- π -|)! _| - |-π-| = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt (9-to-the-(3!-to-the-3)) _| { that is, 196 = 2-to-the-22nd root of 9^(6^3)) rounded down } = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 164! _| {the rounded-down 128th-root of 164!; since we can make 164 with four π, this gives us another way to make 196 with four π} 197 = 200 - 3 = |_ sqrt 8! _| - 3 = |_ sqrt (|- π -| + |- π -|)! _| - |_π_| = |- sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt (9-to-the-(3!-to-the-3)) -| { that is, 197 = 2-to-the-22nd root of 9^(6^3)) rounded up } 198 = 99 + 99 = 200 - 2 = |_ sqrt 8! _| - 2 = |_ sqrt (|- π -| + |- π -|)! _| - |- sqrt π -| 199 = 200 - 1 = |_ sqrt 8! _| - 1 = |_ sqrt (|- π -| + |- π -|)! _| - |_ sqrt π _| 200 = |_ sqrt 8! _| {and, as we can near the top, there are several ways to make 8 with four π, hence we can make 200 with four π} {but we can make 8 with two π, so we can make 200 with two π}: = |_ sqrt (|- π -| + |- π -|)! _| {with two π} 201 = 200 + 1 = |_ sqrt 8! _| + 1 = |_ sqrt (|- π -| + |- π -|)! _| + |_ sqrt π _| 202 = 200 + 2 = |_ sqrt 8! _| + 2 = |_ sqrt (|- π -| + |- π -|)! _| + |- sqrt π -| = |- (sqrt(9!))/9 -| x (sqrt 9) + |_ sqrt sqrt 9 _| [ceiling then floor] 203 = 200 + 3 = |_ sqrt 8! _| + 3 = |_ sqrt (|- π -| + |- π -|)! _| + |_π_| = ( |- (sqrt(9!))/9 -| x 3 ) + |- sqrt 3 -| [ceiling then ceiling] 204 = |_ 12^(π-1) _| = |_ (|-π-| x |_π_|)^(π-1) _| = 200 + 4 = |_ sqrt 8! _| + 4 = |_ sqrt (|- π -| + |- π -|)! _| + |-π-| = ( |_ (sqrt(9!))/9 _| x sqrt 9 ) + (sqrt 9)! = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 165! _| {the rounded-down 128th-root of 165!; since we can make 165 with four π, this gives us another way to make 204 with four π} 205 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {95!} _| {that is, 205 = the 64th root of 95, so insert an existing four π solution for 95 here, rounded down} 206 = 200 + 6 = |_ sqrt 8! _| + 6 = |_ sqrt (|- π -| + |- π -|)! _| + |_π_|! = |- sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt (9-to-the-((sqrt 9)!-to-the-((sqrt 9)))) -| + 9 [ceiling] { that is, 206 = 2-to-the-22nd root of 9^(6^3)) rounded up, + 9 } 207 = 200 + 6 = |_ sqrt 8! _| + 7 = |_ sqrt (|- π -| + |- π -|)! _| + |_π_| + |-π-| = (3! x 3! x 3!) - 9 208 = 200 + 8 = |_ sqrt 8! _| + 8 = |_ sqrt (|- π -| + |- π -|)! _| + |-π-| + |-π-| 209 = 200 + 9 = |_ sqrt 8! _| + 9 = |_ sqrt (|- π -| + |- π -|)! _| + (|_&pi x π_|) = |_ sqrt (9 - |_ sqrt sqrt 9 _| _|)! _| + (9/sqrt 9) = |_ sqrt (9 - |_ sqrt sqrt 9 _| _|)! _| + 9 {that makes 209 with 3 nines, see 200 for details} 210 = (3! x 3! x 3!) - 3! = (|_π_|! x |_π_|! x |_π_|!) - |_π_|! 211 = (3! x 3! x 3!) - 5 = (|_π_|! x |_π_|! x |_π_|!) - |_ sqrt |_ sqrt (|_ π _| !)! _| _| = |_ sqrt (9 - |_ sqrt sqrt 9 _| _|)! _| + 9 + |- sqrt sqrt 9 -| {fancy way of saying 200 + 9 + 2 ... see 200 for details} 212 = (3! x 3! x 3!) - 4 = (|_π_|! x |_π_|! x |_π_|!) - |-π-| = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 166! _| {the rounded-down 128th-root of 166!; since we can make 166 with four π, this gives us another way to make 212 with four π} 213 = (|_π_|! x |_π_|! x |_π_|!) - |_π_| 214 = (|_π_|! x |_π_|! x |_π_|!) - |- sqrt π -| 215 = (|_π_|! x |_π_|! x |_π_|!) - |_ sqrt sqrt 9 _| 216 = 3! ^ 3 = (|_π_|! ^ |_π_|) {with two π} = (|_π_|! x |_π_|! x |_π_|!) x |_ sqrt sqrt 9 _| 217 = (|_π_|! x |_π_|! x |_π_|!) + |_ sqrt π _| 218 = (|_π_|! x |_π_|! x |_π_|!) + |- sqrt π -| 219 = (|_π_|! x |_π_|! x |_π_|!) + |_π_| 220 = |_ π ^ (|_π_| x π) _| {uses three π} {that is, rounded down pi to the power of 3π/2} = |_ sqrt sqrt sqrt sqrt sqrt sqrt {96!} _| {that is, 220 = the 64th root of 96!, so insert an existing four π solution for 96 here, rounded down} 221 = |- π ^ (|_π_| x π) -| {uses three π} {that is, rounded up pi to the power of 3π/2} = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 167! _| {the rounded-down 128th-root of 167!; since we can make 167 with four nines, this gives us another way to make 221 with four nines} 222 = (3! x 3! x 3!) + 3! = (|_π_|! x |_π_|! x |_π_|!) + |_π_|! 223 = 273 - 50 = (3^5) - 50 = |_π_| ^ |_ sqrt |_ sqrt (|_ π _| !)! _| _| - {50 with one π} above gives us 223 with three π} = ( |_ sqrt ((sqrt 9)!)! _| x 9 ) - 9 - |- sqrt sqrt 9 -| [floor then ceiling] 224 = 216 + 8 = (3! ^ 3) + 4 + 4 = (|_π_|! ^ |_π_|) + |-π-| + |-π-| = ( |_ sqrt (3!)! _| x 9 ) - 9 - |_ sqrt 3 _| 225 = (3! ^ 3) + 9 = (|_π_|! ^ |_π_|) + |_ π x π _| = (3! x 3! x 3!) + 9 226 = 273 - 47 = (3^5) - 46 = (|_π_| ^ |_ sqrt |_ sqrt (|_ π _| !)! _| _|) - |- sqrt (|_π_|!! x |_π_|) -| {gives us 226 with three π} 227 = 273 - 46 = (3^5) - 46 = (|_π_| ^ |_ sqrt |_ sqrt (|_ π _| !)! _| _|) - |_ sqrt (|_π_|!! x |_π_|) _| {gives us 227 with three π} 228 = 2 x 114 = 2 x (5! - 6) = 2 x (5! - 3!) = |- sqrt π -| x (|- sqrt (|_ sqrt |_ sqrt (|_ π _| !)! _| _|)! - |_π_|!) {gives us 228 with three π} 229 = {228 with three π} + |_ sqrt π _| 230 = 5 x 46 = {5 with one π} x {46 with two π} {=230 with 3 π} = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 168! _| {the rounded-down 128th-root of 168!; since we can make 168 with four π, this gives us another way to make 230 with four π} 231 = {230 with 3 π} + 1 = {230 with 3 π} + |_ sqrt π _| = ( |_ sqrt ((sqrt 9)!)! _| x 9 ) - (9/sqrt 9) 232 = |_ 5 x sqrt (6!x3) _| = |_ {5 with one π} x |_ sqrt (|_π_|!! x |_π_|) _| _| {with three π} {because 6!x3 = 720x3 = 2160, and sqrt 2160 = 46.47 so 5 sqrt 2160 = 232.379...} 233 = |- 5 x sqrt (6!x3) -| = |_ {5 with one π} x |_ sqrt (|_π_|!! x |_π_|) _| _| {with three π} {because 6!x3 = 720x3 = 2160, and sqrt 2160 = 46.47, so 5 sqrt 2160 = 232.379...} 234 = 26 x 9 = |_ sqrt |_π_|!)! _| x |_ π x π _| {=234 with 3 π} 235 = 5 x 47 = {5 with one π} x {47 with two π} {=235 with 3 π} = |_ sqrt sqrt sqrt sqrt sqrt sqrt {50! x 63!} _| {since we can make each of 50 and 63 with one π, we have 235 with two π}} 236 = {234 with 3 π} + 2 = {234 with 3 π} + |- sqrt π -| = |_ sqrt sqrt sqrt sqrt sqrt sqrt {97!} _| {that is, 236 = the 64th root of 97!, and 97 can be made with two π hence 236 can be made with two π} 237 = {234 with 3 π} + 3 = {234 with 3 π} + |_π_| 238 = {234 with 3 π} + 4 = {234 with 3 π} + |-π-| 239 = ??? = |_ sqrt(sqrt((9^9))) _| + 99 = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 169! _| {the rounded-down 128th-root of 169!; since we can make 169 with four π, this gives us another way to make 239 with four π} 240 = 720/3 = 6!/3 = 3!!/3 = |_π_|!!/|_π_| {with just two π} 241 = 240 + 1 = (720/3) + 1 = (6!/3) + 1 = (3!!/3) + 1 = = |_π_|!!/|_π_| + |_ sqrt π _| {with just three π} 242 = 240 + 2 = (720/3) + 2 = (6!/3) + 2 = (3!!/3) + 2 = = |_π_|!!/|_π_| + |- sqrt π -| {with just three π} 243 = 3^5 = |_π_| ^ |_ sqrt |_ sqrt (|_ π _| !)! _| _| {with two π} = |_ π ^ (24/5) _| {and 24 = 4!, 5 can be made with one π} 244 = (3^5) + 1 = (|_π_| ^ |_ sqrt |_ sqrt (|_ π _| !)! _| _| ) + |_ sqrt π _| {with three π} = |_ sqrt sqrt sqrt sqrt sqrt sqrt {43! x 70!} _| {since we can make each of 43 and 70 with one π, we have 244 with two π} 245 = (3^5) + 2 = (|_π_| ^ |_ sqrt |_ sqrt (|_ π _| !)! _| _| ) + |- sqrt π -| {with three π} 246 = (3^5) + 3 = (|_π_| ^ |_ sqrt |_ sqrt (|_ π _| !)! _| _| ) + |_π_| {with three π} 247 = (3^5) + 4 = (|_π_| ^ |_ sqrt |_ sqrt (|_ π _| !)! _| _| ) + |-π-| {with three π} 248 = (3^5) + 5 = (|_π_| ^ |_ sqrt |_ sqrt (|_ π _| !)! _| _| ) + |_ sqrt |_ sqrt (|_ π _| !)! _| _| 249 = (3^5) + 6 = (3^5) + 3! = (|_π_| ^ |_ sqrt |_ sqrt (|_ π _| !)! _| _| ) + |_π_|! {with three π} = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 170! _| {the rounded-down 128th-root of 170!; since we can make 170 with four π, this gives us another way to make 249 with four π} 250 = 50 x 5 = {50 with one π} x |_ sqrt |_ sqrt (|_ π _| !)! _| _| {= 250 with two π} 251 = {250 with two π} + 1 = {250 with two π} + |_ sqrt π _| {= 251 with three π} 252 = {250 with two π} + 2 = {250 with two π} + |- sqrt π -| {= 252 with three π} 253 = {250 with two π} + 3 = {250 with two π} + |_π_| {= 253 with three π} 254 = {250 with two π} + 4 = {250 with two π} + |-π-| {= 254 with three π} = |_ sqrt sqrt sqrt sqrt sqrt sqrt {98!} _| {that is, 254 = the 64th root of 98!, rounded down, so insert an existing four π solution for 98 here} 255 = {250 with two π} + 5 = {250 with two π} + |_ sqrt |_ sqrt (|_ π _| !)! _| _| {= 255 with three π} = (((sqrt 9)!)! / (sqrt 9)) + 9 + (sqrt 9)! = |_ sqrt sqrt sqrt sqrt sqrt (3!^99) _| {that is, the 32nd root of 6-to-the-99 is 255.50854...} 256 = 2^8 = |- sqrt π -| ^ (|-π-| + |-π-|) {with three π} = {250 with two π} + 6 = {250 with two π} + 3! = {250 with two π} + |_π_|! {= 256 with three π} = |- sqrt sqrt sqrt sqrt sqrt ((sqrt 9)!)-to-the-99 -| 257 = (2^8) + 1 = = (|- sqrt π -| ^ (|-π-| + |-π-|)) + |_ sqrt π _| = |_ sqrt sqrt sqrt sqrt sqrt (3^99 _| + |_ sqrt π _| 258 = (2^8) + 2 = = (|- sqrt π -| ^ (|-π-| + |-π-|)) + |- sqrt π -| 259 = (2^8) + 3 = = (|- sqrt π -| ^ (|-π-| + |-π-|)) + |_π_| = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 171! _| {the rounded-down 128th-root of 171!; since we can make 171 with four π, this gives us another way to make 259 with four π} 260 = (2^8) + 4 = = (|- sqrt π -| ^ (|-π-| + |-π-|)) + |-π-| = |_ ( |_ ( |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt (9^(9^3)) _| ) / (|_ sqrt sqrt 9 _| ) _| 261 = (2^8) + 5 = = (|- sqrt π -| ^ (|-π-| + |-π-|)) + |_ sqrt |_ sqrt (|_ π _| !)! _| _| = |_ sqrt sqrt sqrt sqrt sqrt (3!)^99 _| + 3! = |_ sqrt sqrt sqrt sqrt sqrt sqrt {43! x 71!} _| {since we can make each of 43 and 71 with one π, we have 261 with two π} 262 = (2^8) + 6 = = (|- sqrt π -| ^ (|-π-| + |-π-|)) + |_π_|! = |- sqrt sqrt sqrt sqrt sqrt ((sqrt 9)!)-to-the-99 -| + 3! 263 = 273 - 9 = (3^5) - 9 = ((|_π_|^5) - |_π_|) - |_ |_π_| x |_π_|) = |_ sqrt sqrt sqrt sqrt sqrt sqrt 99! _| - 9 - |_ sqrt sqrt 9 _| 264 = 273 - 8 = (3^5) - 9 = ((|_π_|^5) - |_π_|) - |-π-| - |-π-|) = |_ sqrt sqrt sqrt sqrt sqrt ((sqrt 9)!)-to-the-99 _| + 9 265 = 273 - 7 = (3^5) - 3! - 1 = (|_π_|^5) - |_π_|! - |_ sqrt π _| = |- sqrt sqrt sqrt sqrt sqrt ((sqrt 9)!)-to-the-99 -| + 9 [ceiling] 267 = 273 - 6 = (3^5) - 3! = (|_π_|^5) - |_π_|! {with three π, see 273} 268 = 4^4 + 4!/sqrt(4) = (|-π-|^|-π-|) + (|-π-|!/sqrt |-π-|) = |_ sqrt sqrt sqrt sqrt sqrt sqrt 99! _| + |_ sqrt sqrt 9 _| - (sqrt 9)! 269 = |_ sqrt sqrt sqrt sqrt sqrt sqrt 99! _| - 4 {floor followed by two ceilings} 270 = 273 - 3 = (3^5) - 3 = (|_π_|^5) - |_π_| {with three π, see 273} = |_ sqrt sqrt sqrt sqrt sqrt sqrt 99! _| - (9/sqrt 9) = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 172! _| {the rounded-down 128th-root of 172!; since we can make 172 with four nines, this gives us another way to make 270 with four nines} 271 = |_ sqrt sqrt sqrt sqrt sqrt sqrt 99! _| - 2 272 = 4^4 + 4x4 = (|-π-|^|-π-|) + (|-π-| x |-π-|) = |_ sqrt sqrt sqrt sqrt sqrt sqrt 99! _| - 1 273 = |_ sqrt sqrt sqrt sqrt sqrt sqrt 99! _| {that is, 273 is the 64th root of 99!, rounded down, using just three π} 274 = |_ sqrt sqrt sqrt sqrt sqrt sqrt 99! _| + 1 275 = |_ sqrt sqrt sqrt sqrt sqrt sqrt 99! _| + 2 276 = 273 + 3 = (3^5) + 3 = (|_π_|^5) + |_π_| {with three π, see 273} = 4^4 + 4! - 4 = (|-π-|^|-π-|) + |-π-|! - |-π-| = (((sqrt 9)!)! / (3)) + (3! x 3!) 277 = 273 + 4 = (3^5) + 4 = (|_π_|^5) + |-π-| {with three π, see 273} = |_ sqrt sqrt sqrt sqrt sqrt sqrt 99! _| + 4 278 = |_ 6 ^ π _| = |_ |_π_|! ^ π _| {with two π} = |_ 6 x sqrt (6!x3) _| = |_ |_π_|! x |_ sqrt (|_π_|!! x |_π_|) _| _| {with three π} {because 6!x3 = 720x3 = 2160, and sqrt 2160 = 46.47 so 6 sqrt 2160 = 278.8548...} = 4^4 + 4! - sqrt(4) = (|-π-|^|-π-|) + |-π-|! - |- sqrt π -| 279 = |- 6 ^ π -| = |- |_π_|! ^ π -| {with two π} = |- 6 x sqrt (6!x3) -| = |- |_π_|! x |_ sqrt (|_π_|!! x |_π_|) _| -| {with three π} {because 6!x3 = 720x3 = 2160, and sqrt 2160 = 46.47 so 6 sqrt 2160 = 278.8548...} 280 = 4^4 + 4! = (|-π-|^|-π-|) + |-π-|! {with three π} = 400 - 5! = {to be done} = |_ sqrt(sqrt((9^9))) _| + |_ sqrt(sqrt((9^9))) _| 281 = 4^4 + 4!+ 1 = (|-π-|^|-π-|) + |-π-|! + |_ sqrt π _| = |_ sqrt(sqrt((9-to-the-9))) _| + |- sqrt(sqrt((9-to-the-9))) -| [floor then ceiling] = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 173! _| {the rounded-down 128th-root of 173!; since we can make 173 with four nines, this gives us another way to make 281 with four nines} 282 = 4^4 + 4! + sqrt(4) = (|-π-|^|-π-|) + |-π-|! + |- sqrt π -| = |- sqrt(sqrt((9-to-the-9))) -| + |- sqrt(sqrt((9-to-the-9))) -| [ceiling then ceiling] 283 = |_ 3^(2+π) _| = |_ |- sqrt π -|^(|- sqrt π -| + π) _| = 4^4 + 4! + 3 = (|-π-|^|-π-|) + |-π-|! + |_π_| = |_ sqrt sqrt sqrt sqrt sqrt sqrt 99! _| + 9 + |_ sqrt 3 _| 284 = 4^4 + 4! + 4 = (|-π-|^|-π-|) + |-π-|! + |-π-| = |_ 9 x sqrt 999 _| 285 = {300 with just two π} - (3 x 5) = {300 with just two π} - (|_π_| x |_ sqrt |_ sqrt (|_ π _| !)! _| _|) = |- 9 x sqrt 999 -| 286 = (4! x 4!/sqrt(4)) - sqrt(4) = (|-π-|! x |-π-|! / |- sqrt π -|) - |- sqrt π -| 287 = (4! x 4! - sqrt(4)) / sqrt(4) = (|-π-|! x |-π-|! - |- sqrt π -|) / |- sqrt π -| 288 = {300 with just two π} - 12 = {300 with just two π} - |_π_|! - |_π_|! = 9 x |- sqrt 999 -| 289 = (sqrt(4) + 4! x 4!) / sqrt (4) = (|- sqrt π -| + |-π-|! x |-π-|!) / |- sqrt π -| = {300 made with two nines} - 9 - |- sqrt sqrt 9 -| [ceiling] 290 = (4! x 4! / sqrt(4)) + sqrt(4) = (|-π-|! x |-π-|! / |- sqrt π -|) + |- sqrt π -| = {300 made with two nines} - 9 - |_ sqrt sqrt 9 _| 291 = (4! x 4! / sqrt(4)) + 3 = (|-π-|! x |-π-|! / |- sqrt π -|) + |_π_| 292 = (4! x 4! / sqrt(4)) + 4 = (|-π-|! x |-π-|! / |- sqrt π -|) + |-π-| = |_ sqrt sqrt sqrt sqrt sqrt sqrt {46! x 70!} _| {since we can make each of 46 and 70 with one π, we have 292 with two π} 293 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {100!} _| {that is, 293 = the 64th root of 100!, rounded down, so insert an existing four π solution for 100 here} = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 174! _| {the rounded-down 128th-root of 174!; since we can make 174 with four π, this gives us another way to make 293 with four π} 294 = {300 with just two π} - 6 = {300 with just two π} - |_π_|! 295 = {300 with just two π} - 4 - 1 = {300 with just two π} - |-π-| - |_ sqrt π _| 296 = {300 with just two π} - 4 = {300 with just two π} - |-π-| 297 = 99 x (3!) = {300 with just two π} - 3 = {300 with just two π} - |_π_| 298 = {300 with just two π} - 2 = {300 with just two π} - |- sqrt π -| = |_ sqrt sqrt sqrt sqrt sqrt (9^99) _| / 3 299 = {300 with just two π} - 1 = {300 with just two π} - |_ sqrt π _| 300 = 50 x 3! {note that 50 can be made with just one π, and a whole lot of square roots, factorials, and floors so we can make 300 with just two π} 301 = |- π ^ 5 -| - 6 = |- π ^ (|_ sqrt |_ sqrt (|_ π _| !)! _| _|) -| - |_π_|! {with three π} 302 = |- π ^ 5 -| - 5 = |- π ^ (|_ sqrt |_ sqrt (|_ π _| !)! _| _|) -| - |_ sqrt |_ sqrt (|_ π _| !)! _| _| {with three π} 303 = |- π ^ 5 -| - 4 = |- π ^ (|_ sqrt |_ sqrt (|_ π _| !)! _| _|) -| - |-π-| {with three π} 304 = 4^4 + 4! + 4! = (|-π-|^|-π-|) + |-π-|! + |-π-|! = |- π ^ 5 -| - 3 = |- π ^ (|_ sqrt |_ sqrt (|_ π _| !)! _| _|) -| - |_π_| {with three π} 305 = |- π ^ 5 -| - 2 = |- π ^ (|_ sqrt |_ sqrt (|_ π _| !)! _| _|) -| - |- sqrt π -| {with three π} = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 175! _| {the rounded-down 128th-root of 175!; since we can make 175 with four π, this gives us another way to make 305 with four π} 306 = |- π ^ 5 -| - 1 = |- π ^ (|_ sqrt |_ sqrt (|_ π _| !)! _| _|) -| - |_ sqrt π _| {with three π} = (( |- sqrt(sqrt(9!)) -| + |- sqrt(sqrt(9!)) -| ) x (sqrt 9)!) + (sqrt 9)! 307 = |- π ^ (|_ sqrt |_ sqrt (|_ π _| !)! _| _|) -| {with two π} {that is, 307 is the rounded up π to the 5th power} 308 = |- π ^ 5 -| + 1 = |- π ^ (|_ sqrt |_ sqrt (|_ π _| !)! _| _|) -| + |_ sqrt π _| {with three π} 309 = |- π ^ 5 -| + 2 = |- π ^ (|_ sqrt |_ sqrt (|_ π _| !)! _| _|) -| + |- sqrt π -| {with three π} = 103 x 3, and we see above that we can make 103 with a single π, hence we can make 309 with just two π. 310 = |- π ^ 5 -| + 3 = |- π ^ (|_ sqrt |_ sqrt (|_ π _| !)! _| _|) -| + |_π_| {with three π} 311 = |- π ^ 5 -| + 4 = |- π ^ (|_ sqrt |_ sqrt (|_ π _| !)! _| _|) -| + |-π-| {with three π} = |_ 4^(π + 1) _| = |_ |-π-| ^ (|_π_| + 1) _| {with two π} 312 = |- π ^ 5 -| + 5 = |- π ^ (|_ sqrt |_ sqrt (|_ π _| !)! _| _|) -| + |_ sqrt |_ sqrt (|_ π _| !)! _| _| {with three π} = |_ sqrt sqrt sqrt sqrt sqrt sqrt {46! x 71!} _| {since we can make each of 46 and 71 with one π, we have 312 with two π} 313 = |- π ^ 5 -| + 6 = |- π ^ (|_ sqrt |_ sqrt (|_ π _| !)! _| _|) -| + |_π_|! {with three π} 314 = |- π ^ 5 -| + 7 = |- π ^ (|_ sqrt |_ sqrt (|_ π _| !)! _| _|) -| + |_π_| + |-π-| 315 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {101!} _| {that is, 315 = the 64th root of 101!, rounded down, so insert an existing four π solution for 101 here} 316 = |- π ^ 5 -| + 9 = |- π ^ (|_ sqrt |_ sqrt (|_ π _| !)! _| _|) -| + (|_π_| x |_π_|) 317 = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 176! _| {the rounded-down 128th-root of 176!; since we can make 176 with four nines, this gives us another way to make 317 with four nines} 318 = {300 made with two π} + (3 x 3!) 319 = |- π ^ 5 -| + 12 = |- π ^ (|_ sqrt |_ sqrt (|_ π _| !)! _| _|) -| + (|_π_| x |-π-|) = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt 999! _| {that is, 319 is the 1024th root of 999!, rounded down, using just three nines} 320 = 32 x 10 = (|- π ^ |_ π _| -|) x (|_ π _| x |_ π _|) = {32 with two π} x {10 with a single π} = {320 with three π} = |- sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt 999! -| {that is, 319 is the 1024th root of 999!, rounded up, using just three nines} 321 = {307 made with two π} + 12 322 = |- π ^ 5 -| + (3 x 5) {because we can make a 5 with a single π} 323 = |- π ^ 5 -| + (4 x 4) {because we can make a 5 with a single π} 324 = 3 x 3 x 3! x 3! = |_π_| x |_π_| x |_π_|! x |_π_|! 325 = (4! + 2)! / 4!! / 2 326 = {300 with two π} + {26 with one π} 327 = |- π ^ 5 -| + (5 x 5) {because we can make a 5 with a single π} = {309 made with two nines} + (3 x 6) 328 = {300 with two π} + {26 with one π} + 2 329 = {300 with two π} + {26 with one π} + |_π_| 330 = |_ 15^(π-1) _| = |_ (|_π_| x {5 with one π})^(π-1) _| = {300 with two π} + {26 with one π} + |-π-| 331 = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 177! _| {the rounded-down 128th-root of 177!; since we can make 177 with four π, this gives us another way to make 331 with four π} 332 = {300 with two π} + {26 with one π} + |_π_|! 333 = {307 with two π} + {26 with one π} = 999/3 334 = {300 with two π} + 4! + {10 with one π} 335 = {307 with two π} + {26 with one π} + |-π-| 336 = {300 with two π} + {26 with one π} + {10 with one π} 339 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {102!} _| {that is, 339 = the 64th root of 102!, rounded down, so insert an existing four π solution for 102 here} 340 = {307 with two π} + {26 with one π} + {5 with one π} 341 = {307 with two π} + {26 with one π} + |-π-|! 342 = (7^3) - 1 = ((|_π_| + |-π-|) ^ |_π_|) - |_ sqrt π _| = ((sqrt 9)!)! / |- sqrt sqrt 9 -| - 9 - 9 [ceiling] 343 = 7^3 = (|_π_| + |-π-|) ^ |_π_| {with three &pi} 344 = (7^3) + 1 = ((|_π_| + |-π-|) ^ |_π_|) + |_ sqrt π _| = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 178! _| {the rounded-down 128th-root of 178!; since we can make 178 with four π, this gives us another way to make 344 with four π} 345 = (7^3) + 2 = ((|_π_| + |-π-|) ^ |_π_|) + |- sqrt π -| 346 = (7^3) + 3 = ((|_π_| + |-π-|) ^ |_π_|) + |_π_| = |_ ((3!^9) / 9 _| - 3! 347 = (7^3) + 4 = ((|_π_| + |-π-|) ^ |_π_|) + |-π-| 348 = (6! - 4!)/2 = (|_π_|!! - |-π-|!) / |- sqrt π-| 349 = (7^3) + 6 = ((|_π_| + |-π-|) ^ |_π_|) + |_π_|! 350 = {50 with a single π} x 7 = {50 with a single π} x (|_π_| + |_π_|) {with three π} 351 = ({50 with a single π} x 7) + 1 = {to be done} 352 = ({50 with a single π} x 7) + 2 353 = (7^3) + 10 = ((|_π_| + |-π-|) ^ |_π_|) + {10 with one π} = {50 with a single π} x (|_π_| + |_π_|) + |_π_| 354 = {50 with a single π} x (|_π_| + |_π_|) + |-π-| 355 = ({50 with a single π} x 7) + {5 with a single π} 356 = (6! / 2) - 4 = (|_π_|!! / |-π-|) - |-π-| 357 = (6!/2) - 3 = (|_π_|!! / |- sqrt π -|) - |_π_| = ((sqrt 9)!)! / |- sqrt sqrt 9 -| - (9 / sqrt 9) [ceiling] 358 = (6! - 4)/2 = (|_π_|!! - |-π-|) / |- sqrt π -| = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 179! _| {the rounded-down 128th-root of 179!; since we can make 179 with four π, this gives us another way to make 358 with four π} 359 = |_ π^(2+π) _| = |_ π^(|- sqrt π -| + π) _| = (6! - 2)/2 = (|_π_|!! - |- sqrt π -|) / |- sqrt π -| 360 = 6!/2 = |_π_|!! / |- sqrt π -| 361 = (6! + 2)/2 = (|_π_|!! + |- sqrt π -|) / |- sqrt π -| 362 = (6!/2) + 2 = (|_π_|!! / |- sqrt π -|) + |- sqrt π -| 363 = (6!/2) + 3 = (|_π_|!! / |- sqrt π -|) + |_sqrt π_| 364 = (6!/2) + 4 = (|_π_|!! / |- sqrt π -|) + |-sqrt π-| 365 = (|_π_|!! / |- sqrt π -|) + |_π_|! - |_ sqrt π _| = |_ sqrt sqrt sqrt sqrt sqrt sqrt {103!} _| {that is, 365 = the 64th root of 103!, rounded down, so insert an existing four π solution for 103 here} 366 = (|_π_|!! / |- sqrt π -|) + |_π_|! 367 = (|_π_|!! / |- sqrt π -|) + |_π_|! + |_ sqrt π _| 368 = (|_π_|!! / |- sqrt π -|) + |_π_|! + |-π-| + |-π-| 369 = (|_π_|!! / |- sqrt π -|) + |_π_|! + (|_ π x -π _|) 370 = (|_π_|!! / |- sqrt π -|) + |_π_|! + (|- π x -π -|) = 396 - {26 with one π} = |_ π ^ (31/6 _| = |_ π ^ (|_ π ^ |_ π _| _|)/|_π_|!) _| 371 = |- π ^ (31/6 -| = |- π ^ (|_ π ^ |_ π _| _|)/|_π_|!) -| 372 = {396 with two π} - 24 = {396 with two π} - 4! = |_ sqrt sqrt sqrt sqrt (|_π_|! x |_π_|!)! _| - |- π-|! = |_ sqrt sqrt sqrt sqrt sqrt sqrt {50! x 70!} _| {since we can make each of 70 and 50 with one π, we have 372 with two π}} 373 = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 180! _| {the rounded-down 128th-root of 180!; since we can make 180 with four π, this gives us another way to make 373 with four π} 374 = 400 - {26 with a single π} 375 = 360 + 15 = (6!/2) + (3x5) = = (|_π_|!! / |- sqrt π -|) + (|_π_| x {5 with one π} 376 = {400 with three π} - 24 = {400 with three π} - 4! = ((4! - 4)^2) - 4! = = ((|-π-|! - |-π-|) ^ |- sqrt π -|) - |-π-|! 377 = {307 with two π} + {70 with a single π} 378 = 360 + 18 = (6!/2) + (3x6) = = (6!/2) + 4 = (|_π_|!! / |- sqrt π -|) + (|_π_| x |_π_|!) 379 = {384 with two π} - {5 with one π} 380 = {384 with two π} - |-π-| 381 = {300 made with two nines} + (9x9) = {384 with two π} - |_π_| 382 = {384 with two π} - 2 383 = {384 with two π} - 1 384 = {396 with two π} - 12 = |_ π ^ (26/5) _| {both 26 and 5 can be made with one π, so 384 can be made with two π} = |_ sqrt sqrt sqrt sqrt (|_π_|! x |_π_|!)! _| - (|- π-| x |_π_|) 385 = |- π ^ (26/5) -| {both 26 and 5 can be made with one π, so 384 can be made with two π} 386 = {396 with two π} - 10 = |_ sqrt sqrt sqrt sqrt (|_π_|! x |_π_|!)! _| - (|- π x π -|) = 360 + {26 with a single π} 387 = {396 with two π} - 9 = |_ sqrt sqrt sqrt sqrt (|_π_|! x |_π_|!)! _| - (|_π_| x |_π_|) = ((sqrt 9)!)! / |- sqrt sqrt 9 -| + (9 x (sqrt 9)!) 388 = {396 with two π} - 4 - 4 = |_ sqrt sqrt sqrt sqrt (|_π_|! x |_π_|!)! _| - |-π-| - |-π-| 389 = {396 with two π} - 6 - 1 = |_ sqrt sqrt sqrt sqrt (|_π_|! x |_π_|!)! _| - |_π_|! - |_ sqrt π _| = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 181! _| {the rounded-down 128th-root of 181!; since we can make 181 with four π, this gives us another way to make 389 with four π} 390 = {396 with two π} - 6 = |_ sqrt sqrt sqrt sqrt (|_π_|! x |_π_|!)! _| - |_π_|! 391 = {396 with two π} - 4 - 1 = |_ sqrt sqrt sqrt sqrt (|_π_|! x |_π_|!)! _| - |-π-| - |_ sqrt π _| 392 = {396 with two π} - 4 = |_ sqrt sqrt sqrt sqrt (|_π_|! x |_π_|!)! _| - |-π-| = |_ sqrt sqrt sqrt sqrt sqrt sqrt {104!} _| {that is, 392 = the 64th root of 104!, rounded down, so insert an existing four π solution for 104 here} 393 = {396 with two π} - 3 = |_ sqrt sqrt sqrt sqrt (|_π_|! x |_π_|!)! _| - |_π_| 394 = {396 with two π} - 2 = |_ sqrt sqrt sqrt sqrt (|_π_|! x |_π_|!)! _| - |- sqrt π -| 395 = {396 with two π} - 1 = |_ sqrt sqrt sqrt sqrt (|_π_|! x |_π_|!)! _| - |_ sqrt π _| 396 = |_ sqrt sqrt sqrt sqrt (|_π_|! x |_π_|!)! _| {that is to say. 396 = the 16th root of 36! and that uses just two π} 397 = ( |_ (sqrt(9!))/9 _| x (sqrt 9)!) + |_ sqrt sqrt 9 _| = |_ sqrt sqrt sqrt sqrt (3! + 3!)! _| + (9/9) = 400 - 3 = ((4! - 4)^2) - 3 = ((|-π-|! - |-π-|) ^ |- sqrt π -|) - |_π_| 398 = 400 - 2 = ((4! - 4)^2) - 2 = ((|-π-|! - |-π-|) ^ |- sqrt π -|) - |- sqrt π -| = |_ sqrt sqrt sqrt sqrt sqrt sqrt {50! x 71!} _| {since we can make each of 50 and 71 with one π, we have 398 with two π} 399 = 400 - 1 = ((4! - 4)^2) - 1 = ((|-π-|! - |-π-|) ^ |- sqrt π -|) - |_ sqrt π _| 400 = (4! - 4)^2 = (|-π-|! - |-π-|) ^ |- sqrt π -| {with three π} 401 = 400 + 1 = ((4! - 4)^2) + 1 = ((|-π-|! - |-π-|) ^ |- sqrt π -|) + |_ sqrt π _| 402 = 400 + 2 = = ((|-π-|! - |-π-|) ^ |- sqrt π -|) + |- sqrt π -| 403 = 400 + 3 = = ((|-π-|! - |-π-|) ^ |- sqrt π -|) + |_π_| 404 = 400 + 4 = = ((|-π-|! - |-π-|) ^ |- sqrt π -|) + |-π-| 405 = 400 + 5 = {400 with three π} + {5 with one π} = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 182! _| {the rounded-down 128th-root of 182!; since we can make 182 with four π, this gives us another way to make 405 with four π} 406 = 400 + 6 = ((|-π-|! - |-π-|) ^ |- sqrt π -|) + |_π_|! 407 = {307 with two π} + 100 = {307 with two π} + ({10 with one π} x {10 with one π}) = 400 + 7 = |_ 999 / sqrt((sqrt 9)!) _| 408 = 400 + 8 = 409 = 400 + 9 = 410 = 400 + 10 = 411 = {500 with two π} - 81 = {500 with two π} - 3^4 = {500 with two π} - (|_π_| ^ |-π-|) 412 = 400 + 12 = 413 = {396 with two π} + {17 with two π} 414 = {396 with two π} + 18 = {396 with two π} + (|_π_| x |_π_|!) 415 = ??? 416 = {396 with two π} + 20 = {396 with two π} + |-π-|! - |-π-| 417 = {396 with two π} + 21 = {396 with two π} + |-π-|! - |_π_| 418 = {396 with two π} + 22 = {396 with two π} + |-π-|! - |- sqrt π -| 419 = {396 with two π} + 23 = {396 with two π} + |-π-|! - |_ sqrt π _| 420 = {396 with two π} + 24 = {396 with two π} + |-π-|! 421 = {440 with two π} - 19 = {440 with two π} - |-π-|! + {5 with one π}) 422 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {105!} _| {that is, 422 = the 64th root of 105!, rounded down, so insert an existing four π solution for 105 here} = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 183! _| {the rounded-down 128th-root of 183!; since we can make 183 with four π, this gives us another way to make 422 with four π} 423 = {440 with two π} - {17 with two π} 424 = 400 + 24 = 400 + 4! = {to be done} 425 = {440 with two π} - 15 = {440 with two π} - (|_π_| x {5 with one π}) 426 = 400 + {26 with a single π} 427 = {440 with two π} - {10 with one π} - |_π_| 428 = {440 with two π} - 12 = {440 with two π} - (|_π_| x |-π-|) 429 = {440 with two π} - {10 with one π} - |_ sqrt π _| 430 = {440 with two π} - {10 with one π} 431 = {440 with two π} - 9 = 440 with two π} - |_ π x π _| 432 = {440 with two π} - 8 = {440 with two π} - |-π-| - |-π-| 433 = {440 with two π} - 7 = {440 with two π} - |_π_| - |-π-| 434 = {440 with two π} - |_π_|! 435 = {440 with two π} - {5 with one π} 436 = {440 with two π} - |-π-| 437 = {440 with two π} - |_π_| 438 = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt (9-to-the-(9!)) _| {that is, with 17 sqare roots nested, 9 to the power of 9! = 438.402046.... with just two nines} 439 = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 184! _| {the rounded-down 128th-root of 184!; since we can make 184 with four π, this gives us another way to make 439 with four π} 440 = {44 with a single π} x {10 with a single π} {=440 with two π} 441 = {440 with two π} + |_ sqrt π _| = 21^2 = ((sqrt 9)!)! / |- sqrt sqrt 9 -| + (9x9) 442 = {440 with two π} + |- sqrt π -| 443 = {440 with two π} + |_π_| 444 = {440 with two π} + |-π-| 445 = {440 with two π} + {5 with one π} 446 = {440 with two π} + |_π_|! 447 = {440 with two π} + 7 = {440 with two π} + |_π_| + |-π-| 448 = {500 with two π} - {26 with one π} - {26 with one π} = |_ π^(16/3) _| = |_ π^((|-π-| x |-π-|)/|_π_|) _| {with three π} 449 = |- π^(16/3) -| = |- π^((|-π-| x |-π-|)/|_π_|) -| {with three π} 450 = 9 x {50 with one π} 451 = |_ 7 ^ π _| = |_ (|_π_| + |-π-|) ^ π _| {with three π} 452 = {451 with three π} + 1 453 = {451 with three π} + 2 454 = {451 with three π} + |_π_| = |_ sqrt sqrt sqrt sqrt sqrt sqrt {106!} _| {that is, 454 = the 64th root of 106!, rounded down, so insert an existing four π solution for 106 here} 455 = {451 with three π} + |-π-| = {440 with two π} + 15 = {440 with two π} + (|_π_| x {5 with one π}) 456 = {440 with two π} + 16 = {440 with two π} + (|-π-| x |-π-|) 457 = {440 with two π} + {17 with two π} 458 = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 185! _| {the rounded-down 128th-root of 185!; since we can make 185 with four π, this gives us another way to make 458 with four π} 459 = ((sqrt 9)!)! / |- sqrt sqrt 9 -| + 99 460 = {440 with two π} + |-π-|! - |-π-| 461 = {440 with two π} + |-π-|! - |_π_| 462 = 440 with two π} + |-π-|! - |- sqrt π -| 463 = {440 with two π} + |-π-|! - |_ sqrt π _| 464 = {500 with two π} - {26 with one π} - {10 with one π} = {440 with two π} + |-π-|! 465 = {440 with two π} + |-π-|! + |_ sqrt π _| 466 = {440 with two π} + |-π-|! + |- sqrt π -| 467 = {440 with two π} + |-π-|! + |_π| 468 = {500 with two π} - 26 - 6 = {500 with two π} - {26 with one π} - |_π_|! 469 = {500 with two π} - {26 with one π} - {5 with one π} 470 = {500 with two π} - 26 - 4 = {500 with two π} - {26 with one π} - |-π-| 471 = 942/2 = |_ sqrt sqrt sqrt 24! _|/2 = = |_ sqrt sqrt sqrt |- π -|!! _| / |- sqrt π -| {with two π} = {500 with two π} - 26 - 3 = {500 with two π} - {26 with one π} - |_π_| 472 = {500 with two π} - 26 = {500 with two π} - {26 with one π} - 2 473 = {500 with two π} - 26 = {500 with two π} - {26 with one π} - 1 474 = {500 with two π} - 26 = {500 with two π} - {26 with one π} {474 with three π} 475 = {500 with two π} - 26 = {500 with two π} - {26 with one π} + 1 476 = {500 with two π} - 26 = {500 with two π} - {26 with one π} + 2 477 = {500 with two π} - 26 + 3 = {500 with two π} - {26 with one π} + |_π_| = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 186! _| {the rounded-down 128th-root of 186!; since we can make 186 with four π, this gives us another way to make 477 with four nines} 478 = {500 with two π} - 26 + 4 = {500 with two π} - {26 with one π} + |-π-| 479 = {500 with two π} - 26 + 5 = {500 with two π} - {26 with one π} + {5 with one π} 480 = {500 with two π} - 24 + 4 = {500 with two π} - |-π-|! + 4 481 = {500 with two π} - 24 + 5 = {500 with two π} - |-π-|! + {5 with one π} 482 = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt (9-to-the-(3!!) _| {that is, the 256th root of 9^6! = 256th root of 9^720 = 482.8442... with just three π} 483 = {500 with two π} - {17 with one π} 484 = {500 with two π} - 16 = {500 with two π} - (|-π-| x |-π-|) 485 = {500 with two π} - 15 = {500 with two π} - (|_π_| x {5 with one π}) 486 = {440 with two π} + {46 with one π} 487 = {440 with two π} + {46 with one π} + 1 488 = {500 with two π} - 12 = {500 with two π} - (|-π-| x |_π_|) = |_ sqrt sqrt sqrt sqrt sqrt sqrt {107!} _| {that is, 488 = the 64th root of 107!, rounded down, so insert an existing four π solution for 107 here} 489 = {440 with two π} + {46 with one π} + |_π_| 490 = {500 with two π} - 10 = {500 with two π} - {10 with one π} 491 = {500 with two π} - 9 = {500 with two π} - (|_π_| x |_π_|) 492 = {440 with two π} + {46 with one π} + |_π_|! 493 = {500 with two π} - |_π_| - |-π-| 494 = {500 with two π} - |_π_|! = |_ sqrt 9! _| - 99 - 9 495 = {500 with two π} - {5 with one π} 496 = {500 with two π} - |-π-| = |_ sqrt sqrt sqrt sqrt 37! _| {since we can make 37 with two π} this gives us 496 with two π} 497 = |- sqrt sqrt sqrt sqrt 37! -| {since we can make 37 with two π} this gives us 497 with two π} = {500 with two π} - |_π_| = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 187! _| {the rounded-down 128th-root of 187!; since we can make 187 with four π, this gives us another way to make 497 with four π} 498 = {500 with two π} - 2 499 = {500 with two π} - 1 500 = {50 with one π} x {10 with one π} {=500 with two π} 501 = {500 with two π} + 1 = |_ sqrt 9! _| - 99 - |- sqrt sqrt 9 -| [floor then ceiling] 502 = {500 with two π} + 2 = 602 - 100 = |_ sqrt 9! _| - 100 = |_ sqrt |_ π x π _|! _| - ({10 with one π} x {10 with one π}) 503 = {500 with two π} + |_π_| = |_ sqrt 9! _| - 99 {makes 503 with just three nines} 504 = {500 with two π} + |-π-| = |_ sqrt 9! _| - 99 + |_ sqrt sqrt 9 _| 505 = {500 with two π} + {5 with one π} 506 = {500 with two π} + |_π_|! 507 = {500 with two π} + |_π_| + |-π-| 508 = {500 with two π} + |-π-| + |-π-| 509 = {500 with two π} + |_ π x π _| 510 = {500 with two π} + {10 with one π} 511 = {500 with two π} + {10 with one π} + 1 512 = 2^9 = |- sqrt π -| ^ |_ π x π _| {with three π} = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt (9-to-the-(9-to-the-(sqrt 9))) _| - 9 = |_ sqrt 9! _| - 99 + 9 513 = (2^9) + 1 = (|- sqrt π -| ^ |_ π x π _|) + 1 = |- sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt (9-to-the-(9-to-the-(sqrt 9))) -| - 9 [ceiling] 514 = (2^9) + 2 = (|- sqrt π -| ^ |_ π x π _|) + 2 515 = (2^9) + 3 = (|- sqrt π -| ^ |_ π x π _|) + |_π_| = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt (9-to-the-(9-to-the-3)) _| - 3! 516 = (2^9) + 3 = (|- sqrt π -| ^ |_ π x π _|) + |-π-| = |- sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt (9-to-the-(9-to-the-3)) -| - 3! [ceiling] 517 = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 188! _| {the rounded-down 128th-root of 188!; since we can make 188 with four π, this gives us another way to make 517 with four π} 518 = (2^9) + 6 = (|- sqrt π -| ^ |_ π x π _|) + |_π_|! = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt (9-to-the-(9-to-the-(sqrt 9))) _| - 3 519 = (2^9) + 7 = (|- sqrt π -| ^ |_ π x π _|) + |_π_| + |-π-| = |- sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt (9-to-the-(9-to-the-3)) -| - (sqrt 9) [ceiling] 520 = 400 + 120 = {400 with three π} + 5! = 400 with three π} + {5 with one π}! = 2 x (4 + (4^4)) = |- sqrt π -| x (|-π-| + (|-π-|^|-π-|)) 521 = {602 with two π} - 81 = |_ sqrt 9! _| - 81 = |_ sqrt |_ (π x π) _|! _| - (π x π) = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt (9-to-the-(9^3))) _| {that is, 521 = rounded down 256th root of 9^(9^3))} 522 = 528 with three π} - |_π_|! = |- sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt (9-to-the-(9-to-the-(sqrt 9))) -| [with just three nines] {that is, 522 = rounded up 256th root of 9^(9^3))} 523 = 528 with three π} - {5 with one π} = ( |_ (3!)^9 _| / 3! ) - 3! 524 = 4! x (4! - 2) - 4 = |-π-|! x (|-π-|! - |- sqrt π -|) - |-π-| = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt (9-to-the-(9-to-the-(sqrt 9))) _| + (sqrt 9) 525 = {528 with three π} - |_π_| = |_ sqrt sqrt sqrt sqrt sqrt sqrt {108!} _| {that is, 525 = the 64th root of 108!, rounded down, so insert an existing four pi solution for 108 here} 526 = {528 with three π} - |- sqrt π -| = ( |_ (3!^9 _| / 3! ) - 3 527 = {528 with three π} - |_ sqrt π _| = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt (9-to-the-(9-to-the-(sqrt 9))) _| + (sqrt 9)! 528 = (4! - 2) x 4! = (|-π-|! - |- sqrt π -|) x |-π-|! {with three π} = |- sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt (9-to-the-(9-to-the-(sqrt 9))) -| + (sqrt 9)! [ceiling] 529 = {528 with three π} + |_ sqrt π _| = |_ ((sqrt 9)!)-to-the-9 _| / (sqrt 9)! [uses just three nines] 530 = {528 with three π} + |- sqrt π -| = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt (9-to-the-(9-to-the-(sqrt 9))) _| + 9 531 = {528 with three π} + |_π_| = |- sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt (9-to-the-(9-to-the-(sqrt 9))) -| + 9 [ceiling] 532 = {528 with three π} + |-π-| = ( |_ ((sqrt 9)!)-to-the-9 _| / (sqrt 9)! ) + (sqrt 9) 533 = {528 with three π} + {5 with one π} 534 = {528 with three π} + |_π_|! = |_ sqrt sqrt sqrt sqrt sqrt 63! _| {and, as we see above, 63 can be made with a single π, hence 534 can be made with a single π} 535 = {534 with a single π} + |_ sqrt π _| = ( |_ ((sqrt 9)!)-to-the-9 _| / (sqrt 9)! ) + (sqrt 9)! 536 = 2 * (4^4) + 4! = |- sqrt π -| x (|-π-| ^ |-π-|) + |-π-|! 537 = {534 with a single π} + |_π_| 538 = {528 with three π} + {10 with one π} = {534 with a single π} + |_π_| = ( |_ ((sqrt 9)!)-to-the-9 _| / (sqrt 9)! ) + 9 539 = {534 with a single π} + {5 with a single π} = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 189! _| {the rounded-down 128th-root of 189!; since we can make 189 with four π, this gives us another way to make 539 with four π} 540 = {534 with a single π} + |_π_|! = 36 x 15 = |_ π ^ π _| x 3 x {5 with a single π} 541 = {546 with three π} - {5 with one π} 542 = {602 with two π} - {50 with one π} - {10 with one π} = {546 with three π} - 4 543 = {546 with three π} - 3 544 = {546 with three π} - 2 545 = {546 with three π} - 1 546 = (273 x 2) = (3^5) x 2 = (|_π_| ^ |_ sqrt |_ sqrt (|_ π _| !)! _| _|) x |- sqrt π -| {which makes 546 with three π} = {602 with two π} - {50 with one π} - 6 547 = {602 with two π} - {50 with one π} - {55 with one π} 548 = {602 with two π} - {50 with one π} - 4 549 = {602 with two π} - {50 with one π} - 3 550 = {602 with two π} - {50 with one π} - 2 551 = {602 with two π} - {50 with one π} - 1 552 = (|-π-|! x |-π-|!) - 24 = (|-π-|! x |-π-|!) - |-π-|! = {602 with two π} - {50 with one π} 553 = (|-π-|! x |-π-|!) - 23 554 = (|-π-|! x |-π-|!) - 22 555 = (|-π-|! x |-π-|!) - 21 = 37 x 15 = |- π ^ π -| x 3 x {5 with a single π} 556 = (|-π-|! x |-π-|!) - 20 557 = (|-π-|! x |-π-|!) - 19 558 = (|-π-|! x |-π-|!) - 18 559 = (|-π-|! x |-π-|!) - 17 560 = (|-π-|! x |-π-|!) - 16 561 = (|-π-|! x |-π-|!) - 15 562 = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 190! _| {the rounded-down 128th-root of 190!; since we can make 190 with four π, this gives us another way to make 562 with four π} 563 = (|-π-|! x |-π-|!) - 13 564 = (|-π-|! x |-π-|!) - 12 565 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {109!} _| {that is, 565 = the 64th root of 109!, rounded down, so insert an existing four π solution for 109 here} 566 = (|-π-|! x |-π-|!) - 10 567 = (|-π-|! x |-π-|!) - 9 568 = (|-π-|! x |-π-|!) - 8 569 = (|-π-|! x |-π-|!) - 7 570 = (|-π-|! x |-π-|!) - 6 571 = (|-π-|! x |-π-|!) - 5 572 = (|-π-|! x |-π-|!) - 4 573 = (|-π-|! x |-π-|!) - 3 574 = (|-π-|! x |-π-|!) - 2 575 = (|-π-|! x |-π-|!) - 1 = |_ sqrt 9! _| - ((sqrt 9) x (sqrt 9) x (sqrt 9)) 576 = 24 x 24 = 4! x 4! = |-π-|! x |-π-|! {576 with two π} = 602 - 26 = |_ sqrt 9! _| - 26 = |_ sqrt |_ π x π _|! _| - {26 with one π}) = |_ 999/sqrt sqrt 9 _| 577 = (|-π-|! x |-π-|!) + 1 = |- 999/sqrt sqrt 9 -| 578 = (|-π-|! x |-π-|!) + 2 = 602 - 24 = |_ sqrt 9! _| - 4! = |_ sqrt |_ π x π _|! _| - |-π-|! 579 = (|-π-|! x |-π-|!) + 3 580 = (|-π-|! x |-π-|!) + 4 581 = (|-π-|! x |-π-|!) + 5 582 = (|-π-|! x |-π-|!) + 6 583 = (|-π-|! x |-π-|!) + 7 584 = (|-π-|! x |-π-|!) + 8 = 602 - 18 = |_ sqrt 9! _| - 18 = |_ sqrt |_ π x π _|! _| - (|_π_| x |_π_|!) 585 = (|-π-|! x |-π-|!) + 9 = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 191! _| {the rounded-down 128th-root of 191!; since we can make 191 with four π, this gives us another way to make 585 with four π} 586 = (|-π-|! x |-π-|!) + 10 = 602 - 16 = |_ sqrt 9! _| - 16 = |_ sqrt |_ π x π _|! _| - (|- π x |-π-|) 587 = 602 - 15 = |_ sqrt 9! _| - 15 = |_ sqrt |_ π x π _|! _| - (3x5) 588 = 602 - 14 = |_ sqrt 9! _| - 14 = |_ sqrt |_ π x π _|! _| - 14 589 = 602 - 12 = |_ sqrt 9! _| - 13 = |_ sqrt |_ π x π _|! _| - 13 590 = 602 - 12 = |_ sqrt 9! _| - 12 = |_ sqrt |_ π x π _|! _| - (|_ π x |-π-|) 591 = 602 - 10 = |_ sqrt 9! _| - 1 = |_ sqrt |_ π x π _|! _| - 11 592 = 602 - 10 = |_ sqrt 9! _| - 10 = |_ sqrt |_ π x π _|! _| - {10 with one π} 593 = (|-π-|! x |-π-|!) + 11 = |_ sqrt 9! _| - 9 = |_ sqrt |_ π x π _|! _| - |_ π x π _| 594 = (|-π-|! x |-π-|!) + 12 = 99 x 6 = ( |_ (sqrt(9!))/9 _| x 9 ) [66x9 with just 3 nines], so, next: 595 = (|-π-|! x |-π-|!) + 13 = ( |_ (sqrt(9!))/9 _| x 9 ) + |_ sqrt 3 _| 596 = (|-π-|! x |-π-|!) + 14 = |_ sqrt 9! _| - 3! {uses three π} 597 = (|-π-|! x |-π-|!) + 15 = ( |_ (sqrt(9!))/9 _| x 9 ) + 3 598 = (|-π-|! x |-π-|!) + 16 = |_ sqrt 9! _| - 4 599 = (|-π-|! x |-π-|!) + 17 = |_ sqrt 9! _| - 3 = |_ sqrt |_ π x π _|! _| - |_π_| = |_ sqrt 9! _| - 3 600 = 6 x 10 x 10 = |_π_|! x {10 using only one π} x {10 using only one π} {giving us 600 with three π} = {602 with two π} = |_ sqrt |_ π x π _|! _| - 2 = (|-π-|! x |-π-|!) + 18 = ( |_ (sqrt(9!))/9 _| x 9 ) + (sqrt 9)! |_ sqrt 9! _| - |_ sqrt sqrt sqrt sqrt (9!) _| 601 = {600 with three π} + 1 = {602 with two π} - 1 = |_ sqrt 9! _| - 1 = |_ sqrt |_ π x π _|! _| = |_ sqrt 9! _| + |_ sqrt sqrt 9 _| + |_ sqrt sqrt 9 _| - sqrt 9 602 = |_ sqrt 9! _| {uses just one nine hence 602 with two π} = |_ sqrt |_ π x π _|! _| 603 = {602 with two π} + 1 = |_ sqrt |_ π x π _|! _| + 1 = ( |_ (sqrt(9!))/9 _| x 9 ) + 9 604 = {602 with two π} + 2 = |_ sqrt |_ π x π _|! _| + 2 = |_ 273^(π-2) _| = |_ (3^5)^(π-2) _| 605 = |_ sqrt 9! _| + 3 606 = |_ sqrt 9! _| + 4 607 = |_ sqrt 9! _| + 5 608 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {110!} _| {that is, 608 = the 64th root of 110!, rounded down, so insert an existing four π solution for 110 here} = |_ sqrt 9! _| + 3! 609 = |_ sqrt 9! _| + 7 610 = |_ sqrt 9! _| + 8 = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 192! _| {the rounded-down 128th-root of 192!; since we can make 192 with four π, this gives us another way to make 610 with four π} 611 = |_ sqrt 9! _| + 9 612 = |_ sqrt 9! _| + 10 613 = |_ sqrt 9! _| + 11 614 = |_ sqrt 9! _| + 12 615 = |_ sqrt 9! _| + 13 616 = |_ sqrt 9! _| + 14 617 = |_ sqrt 9! _| + 15 618 = |_ sqrt 9! _| + 16 = 103 x 3!, and we see above that we can make 103 with a single π, hence we can make 618 with just two π} 619 = {618 with two π} + 1 620 = {618 with two π} + 2 = 720 - 100 = 6! - 100 = 3!! - (10 x 10) = ((sqrt 9)!)! - 99 - |_ sqrt sqrt 9 _| 621 = {618 with two π} + 3 622 = {618 with two π} + 4 = ((sqrt 9)!)! - 99 + |_ sqrt sqrt 9 _| 623 = {618 with two π} + 5 624 = {618 with two π} + 3! = ((sqrt 9)!)! - 99 + sqrt 9 625 = 5^4 = {5 with a single π} ^ |-π-| {625 with two π} 626 = {625 with two π} + 1 627 = {625 with two π} + 2 628 = {625 with two π} + |_π_| 629 = {625 with two π} + |-π-| 630 = {625 with two π} + {5 with one π} 631 = {625 with two π} + |_π_| 632 = {625 with two π} + |_π_| + |-π-| 633 = {625 with two π} + |-π-| + |-π-| 634 = {625 with two π} + |_ π x π _| 635 = {625 with two π} + {10 with one π} = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 193! _| {the rounded-down 128th-root of 193!; since we can make 193 with four π, this gives us another way to make 635 with four π} 636 = {625 with two π} + {10 with one π} + 1 637 = {625 with two π} + {10 with one π} + 2 638 = {625 with two π} + {10 with one π} + |_π_| 639 = {625 with two π} + {10 with one π} + |-π-| 640 = {625 with two π} + {10 with one π} + {5 with one π} 641 = {625 with two π} + {10 with one π} + |_π_|! 642 = |_π_|!! - (|_π_|^|-π-|) + |_π_| 643 = {625 with two π} + (|_π_| x |-π-|) 644 = ((4! + 2)!/(4!!)) - 3! = ((|-π-|! + |_ sqrt π _|)!/(|-π-|!!)) - |_π_|! 645 = |_π_|!! - (|_π_|^|-π-|) + |_π_|! 646 = ((4! + 2)!/(4!!)) - 4 = ((|-π-|! + |_ sqrt π _|)!/(|-π-|!!)) - |-π-| 647 = {648 with three π} - 1 648 = (3! x 3! x 3!) x 3 = (|_π_| ^ |_π_|) x |_π_| {=648 with three π} 649 = {648 with three π} + |_ sqrt π _| 650 = {648 with three π} + |- sqrt π -| 651 = {648 with three π} + |_π_| = {625 with two π} + {26 with one π} {=651 with three π} 652 = {648 with three π} + |-π-| 653 = {648 with three π} + {5 with one π} 654 = {648 with three π} + |_π_|! = {625 with two π} + {5 with one π} + |-π-|! 655 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {111!} _| {that is, 655 = the 64th root of 111!, rounded down, so insert an existing four π solution for 111 here} 656 = {625 with two π} + {26 with one π} + {5 with one π} 657 = {625 with two π} + {26 with one π} + |_π_|! 658 = {648 with three π} + {10 with one π} 659 = {665 with three π} - 6 = |_ sqrt 9! _| + {63 with one π} - |_π_|! 660 = {665 with three π} - {5 with one π} = |_ sqrt 9! _| + {63 with one π} - |-π-| 661 = {665 with three π} - 4 = |_ sqrt 9! _| + {63 with one π} - |-π-| 662 = {665 with three π} - 3 = |_ sqrt 9! _| + {63 with one π} - |_π_| = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 194! _| {the rounded-down 128th-root of 194!; since we can make 194 with four π, this gives us another way to make 662 with four π} 663 = {665 with three π} - 2 = |_ sqrt 9! _| + {63 with one π} - |- sqrt π -| 664 = {665 with three π} - 1 = |_ sqrt 9! _| + {63 with one π} - |_ sqrt π _| 665 = {602 with two π} + {63 with one π} = |_ sqrt 9! _| + {63 with one π} {=665 with three π} 666 = {665 with three π} + 1 = |_ sqrt 9! _| + {63 with one π} + |_ sqrt π _| = {602 with two π} + {64 with one π} {666 with three π} 667 = {665 with three π} + 2 = |_ sqrt 9! _| + {63 with one π} + |- sqrt π -| 668 = {665 with three π} + 3 = |_ sqrt 9! _| + {63 with one π} + |_ π_| 669 = ({67 with two π} x {10 with one π}) - |_ sqrt π _| 670 = {67 with two π} x {10 with one π} 671 = ({67 with two π} x {10 with one π}) + |_ sqrt π _| 672 = |_ 300^(π-2) _| = |_ (50 x 3!)^(π-2) _| = ({67 with two π} x {10 with one π}) + |- sqrt π -| 673 = ({67 with two π} x {10 with one π}) + |_π_| 674 = ({67 with two π} x {10 with one π}) + |-π-| 675 = ({67 with two π} x {10 with one π}) + {5 with one π} = {665 with three π} + 10 = |_ sqrt 9! _| + {63 with one π} + {10 with one π} 676 = ({67 with two π} x {10 with one π}) + |_π_|! 677 = {602 with two π} + {70 with one π} + {5 with one π} 678 = {602 with two π} + {70 with one π} + |_π_|! 679 = {618 with two π} + {70 with one π} + |_ sqrt π _| 680 = {618 with two π} + {70 with one π} + |- sqrt π -| 681 = |_ 8 ^ π _| - 6 = |_ (|-π-| + |-π-|) ^ π _| - |_π_|! 682 = |_ 8 ^ π _| - 5 = |_ (|-π-| + |-π-|) ^ π _| - {5 with one π} 683 = |_ 8 ^ π _| - 4 = |_ (|-π-| + |-π-|) ^ π _| - |-π-| 684 = |_ 8 ^ π _| - 3 = |_ (|-π-| + |-π-|) ^ π _| - |_π_| 685 = |_ 8 ^ π _| - 2 = |_ (|-π-| + |-π-|) ^ π _| - |- sqrt π -| 686 = |_ 8 ^ π _| - 1 = |_ (|-π-| + |-π-|) ^ π _| - |_ sqrt π _| 687 = |_ 8 ^ π _| = |_ (|-π-| + |-π-|) ^ π _| {with three π} 688 = |- 8 ^ π -| = |- (|-π-| + |-π-|) ^ π -| {with three π} = {687 with three π} + 1 689 = {688 with three π} + 1 = {687 with three π} + 2 690 = {687 with three π} + |_π_| = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 195! _| {the rounded-down 128th-root of 195!; since we can make 195 with four π, this gives us another way to make 690 with four π} 691 = {687 with three π} + |-π-| 692 = {687 with three π} + {5 with one pi} 693 = {687 with three π} + |_π_|! 694 = 700 - 6 = ({70 with one π} x {10 with one π}) - |_π_|! 695 = |_ 309^(π-2) _| = |_ (103 x 3)^(π-2) _| = 700 - 5 = ({70 with one π} x {10 with one π}) - {5 with one π} 696 = 700 - 4 = ({70 with one π} x {10 with one π}) - |-π-| 697 = 700 - 3 = ({70 with one π} x {10 with one π}) - |_π_| 698 = 700 - 2 = ({70 with one π} x {10 with one π}) - |- sqrt π -| 699 = 700 - 1 = ({70 with one π} x {10 with one π}) - |_ sqrt π _| 700 = {70 with one π} x {10 with one π} 701 = 700 + 1 = ({70 with one π} x {10 with one π}) + |_ sqrt π _| 702 = {602 with two π} + ({10 with one π} x {10 with one π}) 703 = 700 + 3 = ({70 with one π} x {10 with one π}) + |_π_| 704 = 700 + 4 = ({70 with one π} x {10 with one π}) + |-π-| 705 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {112!} _| {that is, 705 = the 64th root of 112!, rounded down, so insert an existing four π solution for 112 here} 706 = 700 + 6 = ({70 with one π} x {10 with one π}) + |_π_|! = ({71 with a single π} x {10 with a single π}) - |-π-| 707 = ({71 with a single π} x {10 with a single π}) - |_ π_| 708 = ({71 with a single π} x {10 with a single π}) - |- sqrt π -| 709 = ({71 with a single π} x {10 with a single π}) - |_ sqrt π _| 710 = {71 with a single π} x {10 with a single π} {=710 with two π} 711 = 720 - 9 = |_π_|!! - |_π_|! - |_π_| 712 = 720 - 8 = |_π_|!! - |-π-| - |-π-| 713 = 720 - 7 = |_π_|!! - |_π_| - |-π-| 714 = 720 - 6 = |_π_|!! - |_π_|! 715 = 720 - 5 = |_π_|!! - {5 with one π} 716 = 720 - 4 = |_π_|!! - |-π-| 717 = 720 - 3 = |_π_|!! - |_π_| 718 = 720 - 2 = |_π_|!! - |- sqrt π -| 719 = 720 - 1 = |_π_|!! - |_ sqrt π _| 720 = (3!)! = |_π_|!! {720 with just one π because (3!)! = 6! = 720} = (9 x (9x9)) - 9 721 = 720 + 1 = |_π_|!! + |_ sqrt π _| 722 = 720 + 2 = |_π_|!! + |- sqrt π -| 723 = 720 + 3 = |_π_|!! + |_π_| 724 = 720 + 4 = |_π_|!! + |-π-| 725 = 720 + 5 = |_π_|!! + {5 with one π} = (3^6) - 4 = (|_π_| ^ |_π_|!) - |-π-| 726 = 720 + 6 = |_π_|!! + |_π_|! = (3^6) - 3 = (|_π_| ^ |_π_|!) - |_π_| 727 = (3^6) - 2 = (|_π_| ^ |_π_|!) - |_ sqrt π _| 728 = 9^3 - |_ sqrt π _| 729 = 3^6 = |_π_| ^ |_π_|! {with two π} 730 = 729 + 1 = (3^6) + 1 = (|_π_| ^ |_π_|!) + |_ sqrt π _| 731 = 729 + 2 = (3^6) + 2 = (|_π_| ^ |_π_|!) + |- sqrt π -| 732 = 729 + 3 = (3^6) + 3 = (|_π_| ^ |_π_|!) + |_π_| 733 = 729 + 4 = (3^6) + 4 = (|_π_| ^ |_π_|!) + |-π-| 734 = 729 + 5 = (3^6) + 5 = (|_π_| ^ |_π_|!) + {5 with one π} 735 = 729 + 6 = (3^6) + 6 = (|_π_| ^ |_π_|!) + |_π_|! 736 = 729 + 7 = (3^6) + 7 = (|_π_| ^ |_π_|!) + |_π_| + |-π-| 737 = 729 + 8 = (3^6) + 8 = (|_π_| ^ |_π_|!) + |-π-| + |-π-| 738 = 729 + 9 = (3^6) + 9 = (|_π_| ^ |_π_|!) + |_ π x π _| 739 = 729 + 10 = (3^6) + 10 = (|_π_| ^ |_π_|!) + |- π x π -| 740 = 720 + 24 - 4 = |_π_|!! + |-π-|! - |-π-| 741 = 720 + 24 - 3 = |_π_|!! + |-π-|! - |_π_| 742 = 720 + 24 - 2 = |_π_|!! + |-π-|! - |- sqrt π -| 743 = 720 + 24 - 1 = |_π_|!! + |-π-|! - |_ sqrt π _| 744 = 720 + 24 = |_π_|!! + |-π-|! {with two π} 745 = 720 + 24 + 1 = |_π_|!! + |-π-|! + |_ sqrt π _| 746 = 720 + 24 + 2 = |_π_|!! + |-π-|! + |- sqrt π -| 747 = 720 + 24 + 3 = |_π_|!! + |-π-|! + |_π_| = 3^6 + (3 x 6) 748 = 720 + 24 + 4 = |_π_|!! + |-π-|! + |-π-| 749 = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 197! _| {the rounded-down 128th-root of 197!; since we can make 197 with four π, this gives us another way to make 749 with four π} 750 = 720 + 24 + 6 = |_π_|!! + |-π-|! + |_π_|! = 3 x 5 x 50 = |_π_| x {5 with one π} x {5 with one π} 751 = 720 + 24 + 7 = |_π_|!! + |-π-|! + |_π_| + |-π-| 752 = 700 + 26 + 26 = ({70 with one π}x{10 with one π}) + {26 with one π}+{26 with one π} 753 = 729 + 24 = (3^6) + 4! = (|_π_| ^ |_π_|!) + |-π-|! 754 = 720 + 24 + {10 with one π} = |_π_|!! + |-π-|! + |- π x π -| 755 = 729 + 26 = (3^6) + {26 with one π} = (|_π_| ^ |_π_|!) + {26 with one π} 756 = 720 + 24 + 12 = |_π_|!! + |-π-|! + (|-π-| x |_π_|) = 729 + 24 + 3 = (3^6) + 4! + 3 = (|_π_| ^ |_π_|!) + |-π-|! + |_π_| = 729 + 27 = (3^6) + (3^3) = (|_π_| ^ |_π_|!) + (|_π_| ^ |_π_|) 757 = 729 + 24 + 5 = (3^6) + 4! + 5 = (|_π_| ^ |_π_|!) + |-π-|! + {5 with one π} 758 = 720 + 24 + 12 = |_π_|!! + |-π-|! + (|_ π_| x |-π-|) 759 = 729 + 30 = 729 + 24 + 6 = (3^6) + 4! + 6 = (|_π_| ^ |_π_|!) + |-π-|! + |_π_|!! = |_ sqrt sqrt sqrt sqrt sqrt sqrt {113!} _| {that is, 759 = the 64th root of 113!, rounded down, so insert an existing four π solution for 113 here} 760 = 729 + 31 = (|_π_| ^ |_π_|!) + |_ π ^ |_ π _| _| 761 = 729 + 32 = (|_π_| ^ |_π_|!) + |- π ^ |_ π _| -| 762 = 720 + 42 = |_π_|!! + {43 with one π} - |_ sqrt π _| 763 = 720 + 43 = |_π_|!! + {43 with one π} 764 = 720 + 44 = |_π_|!! + {44 with one π} 765 = 720 + 45 = |_π_|!! + {44 with one π} + |_ sqrt π _| 766 = 720 + 46 = |_π_|!! + |_π_|!! + {43 with one π} + |_π_| 767 = 720 + 44 + 3 = |_π_|!! + {44 with one π} + |_&pi_| 768 = 256 x 3 = (2^8) x 3 = |- sqrt π -| ^ (|-π-| + |-π-|) x |_π_| = 720 + 44 + 4 = |_π_|!! + {44 with one π} + |-π-| 769 = 729 + 40 = (|_π_| ^ |_π_|!) + (|-π-| x {10 with one π} 770 = 729 + 44 - 3 = (|_π_| ^ |_π_|!) + {44 with one π} - |_&pi_| 771 = 729 + 43 - 1 = (|_π_| ^ |_π_|!) + {43 with one π} - |_ sqrt π _| 772 = 729 + 43 = (|_π_| ^ |_π_|!) + {43 with one π} 773 = 729 + 44 = (|_π_| ^ |_π_|!) + {44 with one π} 774 = 729 + 44 + 1 = (|_π_| ^ |_π_|!) + {44 with one π} + |_ sqrt π _| 775 = 729 + 46 = (|_π_| ^ |_π_|!) + {46 with one π} = 729 + 44 + 2 = (|_π_| ^ |_π_|!) + {44 with one π} + |- sqrt π -| 776 = 720 + 36 = 720 + (6x6) = |_π_|!! + (|_π_|! x |_π_|!) 777 = 729 + 44 + 4 = (|_π_| ^ |_π_|!) + {44 with one π} + |-&pi-| 778 = 729 + 44 + 5 = (|_π_| ^ |_π_|!) + {44 with one π} + {5 with one π} 779 = 729 + 44 + 6 = (|_π_| ^ |_π_|!) + {44 with one π} + |_π_|! 780 = 720 + 60 = 720 + (6 x 10) = |_π_|!! + (|_π_|! x {10 with one π}) 781 = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 198! _| {the rounded-down 128th-root of 198!; since we can make 198 with four π, this gives us another way to make 781 with four π} 782 = 720 + 63 - 1 = |_π_|!! + {63 with one π} - |_ sqrt π _| 783 = 720 + 63 = |_π_|!! + {63 with one π} 784 = 720 + 64 = |_π_|!! + {64 with one π} = 720 + 63 + 1 = |_π_|!! + {63 with one π} + |_ sqrt π _| = |_ 5^(π + 1) _| = |_ {5 with one π} ^ (|_π_| + 1) _| {with two π} 785 = 720 + 64 + 1 = |_π_|!! + {64 with one π} + |_ sqrt π _| 786 = 720 + 63 + 3 = |_π_|!! + {63 with one π} + |_π_| 787 = 720 + 63 + 4 = |_π_|!! + {63 with one π} + |-π-| 788 = 720 + 64 + 4 = |_π_|!! + {64 with one π} + |-π-| 789 = 720 + 63 + 6 = |_π_|!! + {63 with one π} + |_π_|! 790 = 720 + 64 + 6 = |_π_|!! + {64 with one π} + |_π_|! 791 = 729 + 63 - 1 = (|_π_| ^ |_π_|!) + {63 with one π} - |_ sqrt π _| 792 = 729 + 63 = (|_π_| ^ |_π_|!) + {63 with one π} 793 = 729 + 64 = (|_π_| ^ |_π_|!) + {64 with one π} 794 = 729 + 64 + 1 = (|_π_| ^ |_π_|!) + {64 with one π} + |_ sqrt π _| 795 = 729 + 63 + 3 = (|_π_| ^ |_π_|!) + {63 with one π} + |_π_| 796 = 729 + 64 + 3 = (|_π_| ^ |_π_|!) + {64 with one π} + |_π_| 797 = 729 + 64 + 4 = (|_π_| ^ |_π_|!) + {64 with one π} + |-π-| 798 = 729 + 70 - 1 = (|_π_| ^ |_π_|!) + {70 with one π} 799 = 729 + 70 = (|_π_| ^ |_π_|!) + {70 with one π} - |_ sqrt π _| 800 = 8 x 10 x 10 = (|-π-| + |-π-|) x {10 with one π} x {10 with one π} = 810 - 10 = ((|_π_|^|-π-|) x {10 with one π}) - {10 with one π} = 200 x 4 = |_ sqrt (|- π -| + |- π -|)! _| x |-π-| {with three π} 801 = 720 + 81 = |_π_|!! + (|_ π _| ^ |- π -|) 802 = 720 + 81 + 1 = |_π_|!! + (|_ π _| ^ |- π -|) + |_ sqrt π _| 803 = 720 + 81 + 2 = |_π_|!! + (|_ π _| ^ |- π -|) + |- sqrt π -|| 804 = 810 - 6 = ((|_π_|^|-π-|) x {10 with one π}) - |_π_|! 805 = 810 - 5 = ((|_π_|^|-π-|) x {10 with one π}) - {5 with one π} 806 = 810 - 4 = ((|_π_|^|-π-|) x {10 with one π}) - |-π-| 807 = 810 - 3 = ((|_π_|^|-π-|) x {10 with one π}) - |_π_| 808 = 810 - 2 = ((|_π_|^|-π-|) x {10 with one π}) - |- sqrt π -| 809 = 810 - 1 = ((|_π_|^|-π-|) x {10 with one π}) - |_ sqrt π _| 810 = 729 + 81 = (|_π_| ^ |_π_|!) + |_ π _| ^ |- π -| = 81 x 10 = (3^4) x {10 with one π} = (|_π_|^|-π-|) x {10 with one π} {=810 with three π}} = (|_π_|^|-π-|) x |- π x π -| 811 = 810 + 1 = ((|_π_|^|-π-|) x {10 with one π}) + |_ sqrt π _| 812 = 810 + 2 = ((|_π_|^|-π-|) x {10 with one π}) + |- sqrt π -| 813 = 810 + 3 = ((|_π_|^|-π-|) x {10 with one π}) + |_π_| 814 = 810 + 4 = ((|_π_|^|-π-|) x {10 with one π}) + |-π-| = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 199! _| {the rounded-down 128th-root of 199!; since we can make 199 with four nines, this gives us another way to make 814 with four nines} 815 = 810 + 5 = ((|_π_|^|-π-|) x {10 with one π}) + {5 with one π} 816 = 810 + 6 = ((|_π_|^|-π-|) x {10 with one π}) + |_π_|! = 859 - 43 = {859 with two π} - {43 with one π} 817 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {114!} _| {that is, 817 = the 64th root of 114!, rounded down, so insert an existing four π solution for 114 here} 818 = 859 - 43 + 2 = {859 with two π} - {43 with one π} + |- sqrt π -| 819 = 846 - 27 = {846 with two π} - (|_π_|^|_π_|) 820 = {810 with three π} + {10 with one π} 821 = 846 - 25 = {846 with two π} = ({5 with one π} x {5 with one π}) = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt (9-to-the-(9-to-the-((sqrt 9)!))) _| / 9 { that is, 821 = 2-to-the-17th root of 9^(9^6)) rounded down, / 0 } 822 = 846 - 24 = {846 with two π} - |-π-|! 823 = 859 - 36 = {859 with two π} - (|_π_|! x |_π_|!) = 720 + {103 with one π} = |_π_|!! + {103 with one π} 824 = (200 x 4) + 4! = (|_ sqrt (|- π -| + |- π -|)! _| x |-π-|) + |-π-| 825 = 700 + 125 = ({70 with one π} x {10 with one π}) + ({5 with one π} ^ |_π_|) 826 = 846 - 20 = {846 with two π} - ({5 with one π} x |-π-|) 827 = 859 - 32 = {859 with two π} - (|- sqrt π-| ^ {5 with one π}) 828 = 720 + {103 with one π} 829 = 729 + 100 = (|_π_| ^ |_π_|!) + ({10 with one π} x {10 with one π}) = 859 - 30 = {859 with two π} - (|_π_| x {5 with one π}) 830 = 720 + 103 + 7 = |_π_|!! + {103 with one π} + |_π_| + |-π-| 831 = 846 - 15 = {846 with two π} - ({5 with one π} x |_π_|) 832 = 859 - 27 = {859 with two π} - (|_π_| ^ |_π_|) 833 = 720 + 103 + 10 = |_π_|!! + {103 with one π} + {10 with one π} 834 = 278 x 3 = |_ 6 ^ π _| x 3 = |_ |_π_|! ^ π _| x |_π_| 835 = 846 - 11 = {846 with two π} - {10 with one π} - |_ sqrt π_| 859 - 24 = {859 with two π} - |-π-|! 836 = 846 - 10 = {846 with two π} - {10 with one π} 837 = 846 - 9 = {846 with two π} - |_ π x π _| 838 = 846 - 8 = {846 with two π} - |-π-| - |-π-| 839 = 846 - 7 = {846 with two π} - |_π_| - |-π-| 840 = 846 - 6 = {846 with two π} - |_π_|! 841 = 846 - 5 = {846 with two π} - {5 with one π} 842 = 846 - 4 = {846 with two π} - |-π-| 843 = 846 - 3 = {846 with two π} - |_π_| 844 = 846 - 2 = {846 with two π} - |- sqrt π-| 845 = 846 - 1 = {846 with two π} - |_ sqrt π_| 846 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {63! x 70!} _| {since we can make each of 70 and 63 with one π, we have 846 with two π} 847 = {846 with two π} + |_ sqrt π_| 848 = {859 with two π} - {10 with one π} - |_ sqrt π_| = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 200! _| {the rounded-down 128th-root of 200!; since we can make 200 with four π, this gives us another way to make 848 with four π} 849 = {859 with two π} - {10 with one π} 850 = {859 with two π} - |_ π x π _| 851 = {859 with two π} - |-π-| - |-π-| 852 = {859 with two π} - |_π_| - |-π-| 853 = {859 with two π} - |_π_|! 854 = {859 with two π} - {5 with one π} 855 = {859 with two π} - |-π-| 856 = {859 with two π} - |_π_| 857 = {859 with two π} - |- sqrt π-| 858 = {859 with two π} - |_ sqrt π_|

859

= |_ sqrt sqrt sqrt sqrt sqrt sqrt {24! x 103!} _| {since we can make each of 24=4! and 103 with one π, we have 859 with two π} 860 = 859 + 1 = {859 with two π} + |_ sqrt π_| = |_ 372^(π-2) _| = |_ {372 with two π}^(π-2) _| 861 = 859 + 2 = {859 with two π} + |- sqrt π-| 862 = 859 + 3 = {859 with two π} + |_π_| 863 = 859 + 4 = {859 with two π} + |-π-| 863 = 859 + 5 = {859 with two π} + {5 with one π} 864 = 859 + 6 = {859 with two π} + |_π_|! 865 = 859 + 7 = {859 with two π} + |_π_| + |-π-| 866 = 859 + 8 = {859 with two π} + |-π-| + |-π-| 867 = 859 + 9 = {859 with two π} + |_ π x π _| 868 = 859 + 10 = {859 with two π} + {10 with one π} 869 = 859 + 11 = {859 with two π} + {5 with one π} + |_π_|! 870 = 859 + 12 = {859 with two π} + (|_π_| x |-π-|) 871 = 846 + 25 = {846 with two π} + ({5 with one π} x {5 with one π}) 872 = 846 + 26 = {846 with two π} + {26 with one π} 873 = 846 + 27 = {846 with two π} + (|_π_| ^ |_π_|) 874 = (440 x 2) - 6 = ({440 with two pi} x |- sqrt π-|) - |_π_|! 875 = 905 - 30 = {905 with two π} - ({5 with one π} x |_π_|!) = 859 + 16 = {859 with two π} + (|-π-| x |-π-|) 876 = 846 + 30 = {846 with two π} + ({5 with one π} x |_π_|!) 877 = 880 - 3 = ({440 with two pi} x |- sqrt π-|) - |_π_| 878 = 905 - 27 = {905 with two π} - (|_π_| ^ |_π_|) 879 = 905 - 26 = {905 with two π} - {26 with one π} 880 = 440 x 2 = {440 with two pi} x |- sqrt π-| = |_ sqrt sqrt sqrt sqrt sqrt sqrt {115!} _| {that is, 880 = the 64th root of 115!, rounded down, so insert an existing four π solution for 115 here} 881 = 905 - 24 = {905 with two π} - |-π-|! 882 = 880 + 2 = ({440 with two pi} x |- sqrt π-|) + |- sqrt π-| 883 = 880 + 3 = ({440 with two pi} x |- sqrt π-|) + |_π_| 884 = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 201! _| {the rounded-down 128th-root of 201!; since we can make 201 with four π, this gives us another way to make 884 with four π} 885 = 905 - 20 = {905 with two π} - (|-π-| x {5 with one π}) 886 = 880 + 6 = ({440 with two pi} x |- sqrt π-|) + |_π_|! 887 = 905 - 18 = {905 with two π} - (|_π_| x |_π_|!) 888 = 905 - 17 = {905 with two π} - {17 with two π} 889 = 905 - 16 = {905 with two π} - (|-π-| x |-π-|) 890 = 905 - 15 = {905 with two π} - (|_π_| x {5 with one π}) 891 = 962 - 71 = (|- π ^ (|_ π_|!) -|) - {71 with one π} 892 = 962 - 70 = (|- π ^ (|_ π_|!) -|) - {70 with one π} = |_ sqrt sqrt sqrt sqrt sqrt (9-to-the-99) _| - sqrt 9 893 = 905 - 12 = {905 with two π} - (|_π_| x |-π-|) 894 = 905 - 11 = {905 with two π} - {5 with one π} - |_π_|! 895 = 905 - 10 = {905 with two π} - {10 with one π} 896 = 905 - 9 = {905 with two π} - |_ π x π _| 897 = 905 - 8 = {905 with two π} - |-π-| - |-π-| 898 = 905 - 7 = {905 with two π} - |-π-| - |_π_| 899 = 905 - 6 = {905 with two π} - |_π_|! 900 = {905 with two π} - {5 with one π} 901 = 905 - 4 = {905 with two π} - |-π-| 902 = 905 - 3 = {905 with two π} - |_π_| 903 = |_ 24^(π-1) _| = |_ |-π-|!^(π-1) _| = 905 - 2 = {905 with two π} - |- sqrt π-| 904 = 905 - 1 = {905 with two π} - |_ sqrt π_|

905

= |_ sqrt sqrt sqrt sqrt sqrt sqrt {63! x 71!} _| {since we can make each of 63 and 71 with one π, we have 905 with two π} 906 = 905 + 1 = {905 with two π} + |_ sqrt π_| 907 = 905 + 2 = {905 with two π} + |- sqrt π-| 908 = 905 + 3 = {905 with two π} + |_π_| 909 = 905 + 4 = {905 with two π} + |-π-| 910 = {905 with two π} + {5 with one π} 911 = 905 + 6 = {905 with two π} + |_π_|! 912 = 905 + 7 = {905 with two π} + |-π-| + |_π_| 913 = 905 + 8 = {905 with two π} + |-π-| + |-π-| 914 = 905 + 9 = {905 with two π} + |_ π x π _| 915 = 905 + 10 = {905 with two π} + {10 with one π} 916 = 905 + 11 = {905 with two π} + {5 with one π} + |_π_|! 917 = 905 + 12 = {905 with two π} + (|_π_| x |-π-|) 918 = 950 - 32 = {950 with two π} - (|- sqrt π-| ^ {5 with one π}) 919 = 905 + 14 = 905 + 24 - 10 = {905 with two π} + |-π-|! - {10 with one π} 920 = 905 + 15 = {905 with two π} + (|_π_| x {5 with one π}) 921 = 905 + 16 = {905 with two π} + (|-π-| x |-π-|) 922 = 905 + 17 = {905 with two π} + {17 with two π} = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 202! _| {the rounded-down 128th-root of 2!; since we can make 202 with four π, this gives us another way to make 922 with four π} 923 = 905 + 18 = {905 with two π} + (|_π_| x |_π_|!) 924 = 905 + 24 - 5 = {905 with two π} + |-π-|! - {5 with one π} 925 = 905 + 20 = {905 with two π} + (|-π-| x {5 with one π}) 926 = 905 + 24 - 3 = {905 with two π} + |-π-|! - |_π_| 927 = 103 x 9, and we see above that we can make 103 with a single π, hence we can make 927 with three π as 103 x |_ π x π _|. 928 = 905 + 24 - 1 = {905 with two π} + |-π-|! - |_ sqrt π_| 929 = 905 + 24 = {905 with two π} + |-π-|! 930 = 905 + 26 - 1 = {905 with two π} + {26 with one π} - |_ sqrt π_| 931 = 905 + 26 = {905 with two π} + {26 with one π} 932 = 905 + 26 + 1 = {905 with two π} + {26 with one π} + |_ sqrt π_| 933 = 905 + 26 + 2 = {905 with two π} + {26 with one π} + |- sqrt π-| 934 = 905 + 26 + 3 = {905 with two π} + {26 with one π} + |_π_| 935 = 905 + 30 = {905 with two π} + ({5 with one π} x |_π_|!) 936 = 26 x 36 = {26 with one π} x |_π_|! x |_π_|! 937 = (103 x 9) + 10 = ({103 with one π} x |_ π x π _|) + {10 with one π} 938 = 950 - 12 = {950 with two π} - (|_π_| x |-π-|) 939 = 950 - 11 = {950 with two π} - {5 with one π} - |_π_|! 940 = 950 - 10 = {950 with two π} - {10 with one π} 941 = 950 - 9 = {950 with two π} - |_ π x π _|

942

= |_ sqrt sqrt sqrt 24! _| = |_ sqrt sqrt sqrt |- π -|!! _| {943 is made with a single π} or 942??? 943 = 950 - 7 = {950 with two π} - |-π-| - |_π_| 944 = 950 - 6 = {950 with two π} - |_π_|! 945 = 950 - 5 = {950 with two π} - {5 with one π} 946 = 950 - 4 = {950 with two π} - |-π-| 947 = 950 - 3 = {950 with two π} - |_π_| 948 = 905 + 43 = {905 with two π} + {5 with one π} = |_ sqrt sqrt sqrt sqrt sqrt sqrt {116!} _| {that is, 948 = the 64th root of 116!, rounded down, so insert an existing four π solution for 116 here} 949 = 950 - 1 = {950 with two π} - |_ sqrt π _|

950

= |_ sqrt sqrt sqrt sqrt sqrt sqrt {26! x 103!} _| {since we can make each of 26 and 103 with one π, we have 950 with two π} 951 = 950 + 1 = {950 with two π} + |_ sqrt π _| 952 = 950 + 2 = {950 with two π} + |- sqrt π-| 953 = 950 + 3 = {950 with two π} + |_π_| 954 = 950 + 4 = {950 with two π} + |-π-| 955 = 950 + 5 = {950 with two π} + {5 with one π} 956 = 950 + 6 = {950 with two π} + |_π_|! 957 = 950 + 7 = {950 with two π} + |-π-| + |_π_| 958 = 950 + 8 = {950 with two π} + |-π-| + |-π-| 959 = 950 + 9 = {950 with two π} + |_ π x π _| 960 = 950 + 10 = {950 with two π} + {10 with one π} 961 = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 203! _| {the rounded-down 128th-root of 203!; since we can make 203 with four π, this gives us another way to make 961 with four π}

962

= |- π ^ (|_ π_|!) -| {with two π} {that is, 962 is the rounded up π to the 6th power} 963 = 962 + 1 = {962 with two π} + |_ sqrt π _| 964 = 962 + 2 = {962 with two π} + |- sqrt π-| 965 = 962 + 3 = {962 with two π} + |_π_| 966 = 962 + 4 = {962 with two π} + |-π-| 967 = 962 + 5 = {962 with two π} + {5 with one π} 968 = 962 + 6 = {962 with two π} + |_π_|! 969 = 962 + 7 = {962 with two π} + |-π-| + |_π_| 970 = 962 + 8 = {962 with two π} + |-π-| + |-π-| 971 = 962 + 9 = {962 with two π} + |_ π x π _| 972 = 962 + 10 = {962 with two π} + {10 with one π} 973 = 962 + 11 = {962 with two π} + {5 with one π} + |_π_|! 974 = 962 + 12 = {962 with two π} + (|_π_| x |-π-|) 975 = 1000 - 25 = {1000 with two π} - ({5 with one π} x {5 with one π}) 976 = 1000 - 24 = {1000 with two π} - |-π-|! 977 = 962 + 15 = {962 with two π} + (|_π_| x {5 with one π}) 978 = 962 + 16 = {962 with two π} + (|-π-| x |-π-|) 979 = 962 + 17 = {962 with two π} + {17 with one π} 980 = 962 + 18 = {962 with two π} + (|_π_| x |_π_|!) 981 = 905 + 70 + 6 = {905 with two π} + {70 with one π} + |_π_|! 982 = 1000 - 18 = {1000 with two π} - (|_π_| x |_π_|!) 983 = 1000 - 17 = {1000 with two π} - {17 with one π} 984 = 1000 - 16 = {1000 with two π} - (|-π-| x |-π-|) 985 = 1000 - 15 = {1000 with two π} - (|_π_| x {5 with one π}) 986 = |_ (1989 - 16) / 2 _| = |_ ({1989 with one π} - (|-π-| x |-π-|)) / |- sqrt π -| _| 987 = (1989 - 15) / 2 = ({1989 with one π} - (|_π_| x {5 with one π})) / |- sqrt π -| 988 = 1000 - 12 = {1000 with two π} - (|_π_| x |-π-|) 989 = 1000 - 11 = {1000 with two π} - {5 with one π} - |_π_|! 990 = 1000 - 10 = {1000 with two π} - {10 with one π} 991 = 1000 - 9 = {1000 with two π} - |_ π x π _| 992 = 1000 - 8 = {1000 with two π} - |-π-| - |-π-| 993 = 1000 - 7 = {1000 with two π} - |-π-| + |_π_| 994 = 1000 - 6 = {1000 with two π} - |_π_|! 995 = |_ 9 ^ π _| = |_ |_ π x π_| ^ π _| {with three π} 996 = |- 9 ^ π -| = |_ |- π x π_| ^ π -| {with three π} 997 = 1000 - 3 = {1000 with two π} - |_π_| 998 = 1000 - 2 = {1000 with two π} - |- sqrt π -| 999 = 1000 - 1 = {1000 with two π} - |_ sqrt π _| 1000 = {10 with a single π} ^ |_π_| {with two π} 1001 = 1000 + 1 = {1000 with two π} + |_ sqrt π _| 1002 = 1000 + 2 = {1000 with two π} + |- sqrt π -| = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 204! _| {the rounded-down 128th-root of 204!; since we can make 204 with four π, this gives us another way to make 1002 with four π} 1003 = 1004 = 1005 = 999 + (sqrt 9)! 1006 = 1007 = 1008 = 999 + 9 ... 1021 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {117!} _| {that is, 1021 = the 64th root of 117!, rounded down, so insert an existing four π solution for 117 here} ... 1044 = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 205! _| {the rounded-down 128th-root of 205!; since we can make 205 with four π, this gives us another way to make 1044 with four π} ... 1058 = |_ ((sqrt 9)!)-to-the-9 _| / (sqrt 9) [uses just three nines] 1059 = 1060 = 1061 = ( |_ ((sqrt 9)!)-to-the-9 _| / (sqrt 9) ) + (sqrt 9) 1062 = ( |- ((sqrt 9)!)-to-the-9 -| / (sqrt 9) ) + (sqrt 9) [ceiling] 1063 = 1064 = ( |_ ((sqrt 9)!)-to-the-9 _| / (sqrt 9) ) + (sqrt 9)! 1065 = ( |- ((sqrt 9)!)-to-the-9 -| / (sqrt 9) ) + (sqrt 9)! [ceiling] 1066 = 1067 = ( |_ ((sqrt 9)!)-to-the-9 _| / (sqrt 9) ) + 9 1067 = |_ 9-to-the-9 / (9!) _| [uses just three nines] 1068 = 1069 = 1070 = |_ 9-to-the-9 / (9!) _| + (sqrt 9) 1071 = |- 9-to-the-9 / (9!) -| + (sqrt 9) [ceiling] 1072 = 1073 = |_ 9-to-the-9 / (9!) _| + (sqrt 9)! 1074 = |- 9-to-the-9 / (9!) -| + (sqrt 9)! [ceiling] 1075 = 1076 = |_ 9-to-the-9 / (9!) _| + 9 1077 = |_ 9-to-the-9 / (9!) _| + 9 [ceiling] ... 1087 = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt 6^999 _| {that is, the 256the root of 6^999 = 1087.96185... } 1088 = |- sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt ((sqrt 9)!)-to-the-999 -| [ceiling] 1089 = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 206! _| {the rounded-down 128th-root of 206!; since we can make 206 with four π, this gives us another way to make 1089 with four π} ... 1100 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {118!} _| {that is, 1100 = the 64th root of 118!, rounded down, so insert an existing four π solution for 118 here} ... 1135 = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 207! _| {the rounded-down 128th-root of 207!; since we can make 207 with four π, this gives us another way to make 1135 with four π} ... 1140 = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt (sqrt 9)-to-the-power-((sqrt 9)-to-the-power-(|_sqrt sqrt 9_|- to-the-power-(sqrt 9)))) _| {that is 1140 = rounded down the 2^10th root of 3^(3^(2^3))} ... 1183 = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 208! _| {the rounded-down 128th-root of 208!; since we can make 208 with four π, this gives us another way to make 1183 with four π} 1184 = 1185 = 1186 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {119!} _| {that is, 1186 = the 64th root of 119!, rounded down, so insert an existing four π solution for 119 here} ... 1232 = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt (9-to-the-(9-to-the-((sqrt 9)!))) _| / ((sqrt 9)!) { that is, 1232 = 2-to-the-17th root of 9^(9^6)) rounded down, / 6 } 1233 = 1234 = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 209! _| {the rounded-down 128th-root of 209!; since we can make 209 with four π, this gives us another way to make 1234 with four π} ... 1246 = |_ 4;^(2+π) _| = |_ |-π-|^(|- sqrt π -| + π) _| ... 1278 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {120!} _| {that is, 1278 = the 64th root of 120!, rounded down, so insert the existing two π solution for 120 here} { note that 120 = 5! } ... 1286 = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 210! _| {the rounded-down 128th-root of 210!; since we can make 210 with four π, this gives us another way to make 1286 with four π} ... 1296 = (sqrt 9)! x (sqrt 9)! x (sqrt 9)! x (sqrt 9)! ... 1329 = |_ π ^ (|_π_| x π)_| {that is, rounded down pi to the power of 2π} 1330 = |- π ^ (|_π_| x π) -| {that is, rounded up pi to the power of 2π} ...

1341

= |_ sqrt sqrt sqrt sqrt sqrt 70! _| {and, as we see above, 70 can be made with a single π, hence 1341 can be made with a single π} = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 211! _| {the rounded-down 128th-root of 211!; since we can make 211 with four π, this gives us another way to make 1341 with four π} ... 1377 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {121!} _| {that is, 1377 = the 64th root of 121!, rounded down, so insert an existing four π solution for 121 here} ... 1385 = |_ 10 ^ π _| = |_ {10 with one π} _| {=1385 with two π} 1386 = |- 10 ^ π -| = |- {10 with one π} -| {=1385 with two π} ... 1399 = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 212! _| {the rounded-down 128th-root of 212!; since we can make 212 with four π, this gives us another way to make 1399 with four π} ...

1433

= |_ sqrt sqrt sqrt sqrt sqrt sqrt {70! x 71!} _| {since we can make each of 70 and 71 with one π, we have 1433 with two π} ... 1458 = 9 x (9 x (9+9)) 1459 = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 213! _| {the rounded-down 128th-root of 213!; since we can make 213 with four π, this gives us another way to make 1459 with four π} ... 1485 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {122!} _| {that is, 1485 = the 64th root of 122!, rounded down, so insert an existing four π solution for 122 here} ... 1521 = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 214! _| {the rounded-down 128th-root of 214!; since we can make 214 with four π, this gives us another way to make 1521 with four π} ... 1586 = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 215! _| {the rounded-down 128th-root of 215!; since we can make 215 with four π, this gives us another way to make 1586 with four π} ... 1601 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {123!} _| {that is, 1601 = the 64th root of 123!, rounded down, so insert an existing four π solution for 123 here} ... 1654 = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 216! _| {the rounded-down 128th-root of 216!; since we can make 216 with four π, this gives us another way to make 1654 with four π} ... 1670 = |_ 6^(π + 1) _| = |_ |_π_|! ^ (|_π_| + 1) _| {with two π} ... 1725 = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 217! _| {the rounded-down 128th-root of 217!; since we can make 217 with four nines, this gives us another way to make 1725 with four nines} 1726 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {124!} _| {that is, 1726 = the 64th root of 124!, rounded down, so insert an existing four π solution for 124 here} ... 1782 = 99 x 3 x 6 ... 1800 = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 218! _| {the rounded-down 128th-root of 218!; since we can make 218 with four π, this gives us another way to make 1800 with four π} ... 1827 = |_ 720^(π-2) _| = |_ |_π_|!!^(π-2) _| ... 1853 = |_ 729^(π-2) _| = |_ (|_π_| ^ |_π_|!)^(π-2) _| ... 1861 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {125!} _| {that is, 1861 = the 64th root of 125!, rounded down, so insert an existing four π solution for 125 here} 1862 = 1863 = 1864 = 1865 = 1866 = 1867 = 1868 = 1869 = |_ 11 ^ π _| = |_ ({10 using only one π} + |_ sqrt π _|) ^ π _| {=1869 with three π} 1877 = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 219! _| {the rounded-down 128th-root of 219!; since we can make 219 with four π, this gives us another way to make 1877 with four π} ...

1904

= |_ sqrt 10! _| {but, as we see near the top, we can make 10 with just a single 9, so we can make 1904 with a single 9} {that is, the rounded-down square root of 10! = sqrt 3628800 = 1904.940943... which rounds down to 1904} {that means that we can make all the years in the 20th century with 4 π} ... 1944 = 9 x (sqrt 9)! x (sqrt 9)! x (sqrt 9)! ... 1958 = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 220! _| {the rounded-down 128th-root of 220!; since we can make 220 with four π, this gives us another way to make 1958 with four π} ...

1989

= |_ sqrt sqrt sqrt sqrt 43! _| {since we can make 43 with one π} this gives us 1989 with one π} 1990 = |_ sqrt sqrt sqrt sqrt 43! _| + 1 {since we can make 43 with one π} this gives us 1990 with two π} 1991 = |_ sqrt sqrt sqrt sqrt 43! _| + 2 {since we can make 43 with one π} this gives us 1991 with two π} 1992 = |_ sqrt sqrt sqrt sqrt 43! _| + 3 {since we can make 43 with one π} this gives us 1992 with two π} 1993 = |_ sqrt sqrt sqrt sqrt 43! _| + 4 {since we can make 43 with one π} this gives us 1993 with two π} 1994 = |_ sqrt sqrt sqrt sqrt 43! _| + 5 {since we can make 43 with one π} this gives us 1994 with two π} 1995 = |_ sqrt sqrt sqrt sqrt 43! _| + 3! {since we can make 43 with one π} this gives us 1995 with two π} 1996 = |_ sqrt sqrt sqrt sqrt 43! _| + 7 {since we can make 43 with one π} this gives us 1996 with two π} 1997 = |_ sqrt sqrt sqrt sqrt 43! _| + 8 {since we can make 43 with one π} this gives us 1997 with two π} 1998 = |_ sqrt sqrt sqrt sqrt 43! _| + 9 {since we can make 43 with one π} this gives us 1998 with two π} 1999 = |_ sqrt sqrt sqrt sqrt 43! _| + 10 {since we can make 43 and 10 each with one π} this gives us 1999 with two π} 2000 = 20 x 100 = (41 - 4) x 100 = = (|-π-|! - |-π-|) x {10 with one π} x {10 with one π} 2001 = can you make it with four π? 2002 = can you make it with four π? 2003 = can you make it with four π? 2004 = can you make it with four π? 2007 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {126!} _| {that is, 2007 = the 64th root of 126!, rounded down, so insert an existing four π solution for 126 here} ... 2042 = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 221! _| {the rounded-down 128th-root of 221!; since we can make 221 with four π, this gives us another way to make 2042 with four π} ...

2116

= |_ sqrt sqrt sqrt 26! _| {since we can make 26 with one π, this gives 2116 with one π} ... 2130 = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 222! _| {the rounded-down 128th-root of 222!; since we can make 222 with four π, this gives us another way to make 2130 with four π} ... 2165 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {127!} _| {that is, 2165 = the 64th root of 127!, rounded down, so insert an existing four π solution for 127 here} ...

2187

= 3^7 = |_π_| ^ (|_π_| + |-π-|) {with three π} ... 2222 = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 223! _| {the rounded-down 128th-root of 223!; since we can make 223 with four π, this gives us another way to make 2222 with four π} ... 2318 = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 224! _| {the rounded-down 128th-root of 224!; since we can make 224 with four π, this gives us another way to make 2318 with four π} ... 2336 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {128!} _| {that is, 2336 = the 64th root of 128!, rounded down, so insert an existing four π solution for 128 here} ... 2368 = |_ 999^9 _| 2368 = |- 999^9 -| [ceiling] ... 2418 = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 225! _| {the rounded-down 128th-root of 225!; since we can make 225 with four π, this gives us another way to make 2418 with four π} ... 2456 = |_ 12 ^ π _| = |_ (|_π_| x |-π-|) ^ π _| {=1256 with three π} ... 2465 = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt (9-to-the-(9-to-the-((sqrt 9)!))) _| / (sqrt 9) { that is, 3698 = 2-to-the-17th root of 9^(9^6)) rounded down, / 3 } ... 2520 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {129!} _| {that is, 2520 = the 64th root of 129!, rounded down, so insert an existing four π solution for 129 here} 2521 = 2522 = 2523 = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 226! _| {the rounded-down 128th-root of 226!; since we can make 226 with four π, this gives us another way to make 2523 with four π} ... 2632 = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 227! _| {the rounded-down 128th-root of 227!; since we can make 227 with four π, this gives us another way to make 2632 with four π} ... 2719 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {130!} _| {that is, 2719 = the 64th root of 130!, rounded down, so insert an existing four π solution for 130 here} ... 2746 = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 228! _| {the rounded-down 128th-root of 228!; since we can make 228 with four π, this gives us another way to make 2746 with four π} ... 2865 = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 229! _| {the rounded-down 128th-root of 229!; since we can make 229 with four π, this gives us another way to make 2865 with four π} ... 2935 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {131!} _| {that is, 2935 = the 64th root of 131!, rounded down, so insert an existing four nines solution for 131 here} ... 2990 = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 230! _| {the rounded-down 128th-root of 230!; since we can make 230 with four nines, this gives us another way to make 2990 with four nines} ... 2997 = 999 x 3 ... 3120 = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 231! _| {the rounded-down 128th-root of 231!; since we can make 230 with four π, this gives us another way to make 3120 with four π} ... 3167 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {132!} _| {that is, 3167 = the 64th root of 132!, rounded down, so insert an existing four π solution for 132 here} ... 3173 = |_ 3!^9 _| - 1 3174 = |_ 3!^9 _| = [1296 sqrt 6] [uses just two nines] = |_ ((sqrt 9)!)-to-the-9 _| x (9/9) 3175 = |_ ((sqrt 9)!)-to-the-9 _| + (9/9) ... 3255 = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 232! _| {the rounded-down 128th-root of 232!; since we can make 232 with four π, this gives us another way to make 3255 with four π} ... 3397 = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 233! _| {the rounded-down 128th-root of 233!; since we can make 233 with four π, this gives us another way to make 3397 with four π} ... 3419 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {133!} _| {that is, 3419 = the 64th root of 133!, rounded down, so insert an existing four π solution for 133 here} ... 3545 = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 234! _| {the rounded-down 128th-root of 234!; since we can make 234 with four π, this gives us another way to make 2990 with four π} ... 3691 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {134!} _| {that is, 3691 = the 64th root of 134!, rounded down, so insert an existing four π solution for 134 here} ... 3698 = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt (9-to-the-(9-to-the-((sqrt 9)!))) _| / |- sqrt sqrt 9 -| [floor divided by ceiling] { that is, 3698 = 2-to-the-17th root of 9^(9^6)) rounded down, / 2 } 3699 = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 235! _| {the rounded-down 128th-root of 235!; since we can make 235 with four π, this gives us another way to make 3699 with four π} ... 3861 = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 236! _| {the rounded-down 128th-root of 236!; since we can make 236 with four π, this gives us another way to make 3861 with four π} ... 3924 = |_ 5^(2+π) _| = |_ { 5 with one π}^(|- sqrt π -| + π) _| ... 3985 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {135!} _| {that is, 3985 = the 64th root of 135!, rounded down, so insert an existing four π solution for 135 here} ... 4029 = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 237! _| {the rounded-down 128th-root of 237!; since we can make 237 with four π, this gives us another way to make 4029 with four π} ...

4062

= |_ sqrt sqrt sqrt sqrt 46! _| {since we can make 46 with one π} this gives us 4062 with one π} ... 4205 = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 238! _| {the rounded-down 128th-root of 238!; since we can make 238 with four π, this gives us another way to make 4205 with four π} ... 4303 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {136!} _| {that is, 4303 = the 64th root of 136!, rounded down, so insert an existing four π solution for 136 here} ...

4374

= 3^6 x 3! = (3^(3!)) x 3! = (|_π_| ^ |_π_|!) x |_π_|! ... 4389 = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 239! _| {the rounded-down 128th-root of 239!; since we can make 239 with four π, this gives us another way to make 4389 with four π} ... 4581 = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 240! _| {the rounded-down 128th-root of 240!; since we can make 240 with four π, this gives us another way to make 4581 with four π} ... 4647 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {137!} _| {that is, 4647 = the 64th root of 137!, rounded down, so insert an existing four π solution for 137 here} ... 4782 = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 241! _| {the rounded-down 128th-root of 241!; since we can make 241 with four π, this gives us another way to make 4782 with four π} ... 4991 = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 242! _| {the rounded-down 128th-root of 242!; since we can make 242 with four π, this gives us another way to make 4991 with four π} ... 5019 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {138!} _| {that is, 5019 = the 64th root of 138!, rounded down, so insert an existing four π solution for 138 here} ... 5210 = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 243! _| {the rounded-down 128th-root of 243!; since we can make 243 with four π, this gives us another way to make 5210 with four π} ... 5421 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {139!} _| {that is, 5421 = the 64th root of 139!, rounded down, so insert an existing four π solution for 139 here} ... 5856 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {140!} _| {that is, 5856 = the 64th root of 140!, rounded down, so insert an existing four π solution for 140 here} ... 5994 = 999 x 3! ... 6317 = |_ sqrt 11! _| {but, as we see near the top, we can make 11 with just three nines, so we can make 6317 with three π}: ... 6327 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {141!} _| {that is, 6327 = the 64th root of 141!, rounded down, so insert an existing four π solution for 141 here} ...

6561

= 3^8 ... 6836 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {142!} _| {that is, 6836 = the 64th root of 142!, rounded down, so insert an existing four π solution for 142 here} ... 7388 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {143!} _| {that is, 7388 = the 64th root of 143!, rounded down, so insert an existing four π solution for 143 here} ... 7396 = |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt (9-to-the-(9-to-the-((sqrt 9)!))) _| [with just three nines] { that is, 7396 = 2-to-the-17th root of 9^(9^6)) rounded down } ... 7984 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {144!} _| {that is, 7984 = the 64th root of 144!, rounded down, so insert an existing four π solution for 144 here} ... 8019 = 9 x 9 x 99 ... 8630 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {145!} _| {that is, 8630 = the 64th root of 145!, rounded down, so insert an existing four π solution for 145 here} ... 8991 = 999 x 9 ... 9329 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {146!} _| {that is, 9329 = the 64th root of 146!, rounded down, so insert an existing four π solution for 146 here} ... 9801 = 99 x 99 ... 10,000 = {10 with a single π} ^ |-π-| {with two π} ... 10,085 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {147!} _| {that is, 10,085 = the 64th root of 147!, rounded down, so insert an existing four π solution for 147 here} ... 10,021 = |_ 6^(2+π) _| = |_ |_π_|!^(|- sqrt π -| + π) _| ... 10,904 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {148!} _| {that is, 10,904 = the 64th root of 148!, rounded down, so insert an existing four π solution for 148 here} ... 11,791 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {149!} _| {that is, 11,791 = the 64th root of 149!, rounded down, so insert an existing four π solution for 149 here} ... 12,751 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {150!} _| {that is, 12,751 = the 64th root of 150!, rounded down, so insert an existing four π solution for 150 here} ... 13,791 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {151!} _| {that is, 13,791 = the 64th root of 151!, rounded down, so insert an existing four π solution for 151 here} ... 14,918 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {152!} _| {that is, 14,918 = the 64th root of 152!, rounded down, so insert an existing four nines solution for 152 here} ... 16,138 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {153!} _| {that is, 16,138 = the 64th root of 153!, rounded down, so insert an existing four π solution for 153 here} ... 17,459 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {154!} _| {that is, 17,459 = the 64th root of 154!, rounded down, so insert an existing four nines solution for 154 here} ... 18,890 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {155!} _| {that is, 18,890 = the 64th root of 155!, rounded down, so insert an existing four π solution for 155 here}... ... 19,682 = |_ sqrt(9-to-the-9) _| - (9/9) 19,683 = |_ sqrt(9-to-the-9) _| = 3-to-the-9 = |_ sqrt((sqrt 9)-to-the-9)) _| which uses just two nines, so: 19,683 = |_ sqrt(9-to-the-9) _| x (9/9) 19,684 = |_ sqrt(9-to-the-9) _| + (9/9) ... 20,441 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {156!} _| {that is, 20,441 = the 64th root of 156!, rounded down, so insert an existing four π solution for 156 here} ...

21,886

= |_ sqrt 12! _| {but, as we see near the top, we can make 12 with just two π, so we can make 21,886 with two π}: = |_ sqrt (_|π_| x |-π-|)! _| ... 22,122 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {157!} _| {that is, 22,122 = the 64th root of 157!, rounded down, so insert an existing four π solution for 157 here} ... 23,943 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {158!} _| {that is, 23,943 = the 64th root of 158!, rounded down, so insert an existing four π solution for 158 here} ... 25,916 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {159!} _| {that is, 25,916 = the 64th root of 159!, rounded down, so insert an existing four π solution for 159 here} ... 28,055 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {160!} _| {that is, 28,055 = the 64th root of 160!, rounded down, so insert an existing four π solution for 160 here} ...

29,809

= |_π^9_| = |_π_| ^ (|_π_| x |_π_|} ... 30,374 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {161!} _| {that is, 30,374 = the 64th root of 161!, rounded down, so insert an existing four π solution for 161 here} ... 30,915 = |_ sqrt sqrt 99^9 _| ... 32,887 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {162!} _| {that is, 32,887 = the 64th root of 162!, rounded down, so insert an existing four π solution for 162 here} ... 35,611 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {163!} _| {that is, 35,611 = the 64th root of 163!, rounded down, so insert an existing four π solution for 163 here} ... 38,565 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {164!} _| {that is, 38,565 = the 64th root of 164!, rounded down, so insert an existing four π solution for 164 here} ... 41,768 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {165!} _| {that is, 41,768 = the 64th root of 165!, rounded down, so insert an existing four π solution for 165 here} ... 45,241 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {166!} _| {that is, 45,241 = the 64th root of 166!, rounded down, so insert an existing four π solution for 166 here} ... 49,008 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {167!} _| {that is, 49,008 = the 64th root of 167!, rounded down, so insert an existing four π solution for 167 here} ... 53,093 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {168!} _| {that is, 53,093 = the 64th root of 168!, rounded down, so insert an existing four nines solution for 168 here} ... 57,523 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {169!} _| {that is, 57,523 = the 64th root of 169!, rounded down, so insert an existing four nines solution for 169 here} ... 62,330 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {170!} _| {that is, 62,330 = the 64th root of 170!, rounded down, so insert an existing four nines solution for 170 here} ... 67,544 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {171!} _| {that is, 62,330 = the 64th root of 171!, rounded down, so insert an existing four nines solution for 171 here} ... 73,201 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {172!} _| {that is, 73,201 = the 64th root of 172!, rounded down, so insert an existing four nines solution for 172 here} ... 78,911 = |_ sqrt 13! _| {but, as we see near the top, we can make 13 with four nines, so we can make 78,911 with four nines} ... 79,339 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {173!} _| {that is, 79,339 = the 64th root of 173!, rounded down, so insert an existing four nines solution for 173 here} ... 86,000 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {174!} _| {that is, 86,000 = the 64th root of 174!, rounded down, so insert an existing four nines solution for 174 here} {exact multiple of 1000} ...

93,648

= |_π^10_| and we can make a 10 with a single π ... 93,227 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {175!} _| {that is, 93,227 = the 64th root of 175!, rounded down, so insert an existing four nines solution for 175 here} ... 100,000 = {10 with a single π} ^ {5 with a single π} {with two π} ... 101,072 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {176!} _| {that is, 101,072 = the 64th root of 176!, rounded down, so insert an existing four nines solution for 176 here} ... 109,586 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {177!} _| {that is, 109,586 = the 64th root of 177!, rounded down, so insert an existing four nines solution for 177 here} ... 118,828 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {178!} _| {that is, 118,828 = the 64th root of 178!, rounded down, so insert an existing four nines solution for 178 here} ... 128,860 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {179!} _| {that is, 128,860 = the 64th root of 179!, rounded down, so insert an existing four nines solution for 179 here} ... 139,752 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {180!} _| {that is, 139,752 = the 64th root of 180!, rounded down, so insert an existing four nines solution for 180 here} ... 151,577 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {181!} _| {that is, 151,577 = the 64th root of 181!, rounded down, so insert an existing four nines solution for 181 here} ... 164,417 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {182!} _| {that is, 164,417 = the 64th root of 182!, rounded down, so insert an existing four nines solution for 182 here} ... 178,360 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {183!} _| {that is, 178,360 = the 64th root of 183!, rounded down, so insert an existing four nines solution for 183 here} ... 193,502 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {184!} _| {that is, 193,502 = the 64th root of 184!, rounded down, so insert an existing four nines solution for 184 here} ... 209,948 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {185!} _| {that is, 209,948 = the 64th root of 185!, rounded down, so insert an existing four nines solution for 185 here} ... 227,810 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {186!} _| {that is, 227,810 = the 64th root of 186!, rounded down, so insert an existing four nines solution for 186 here} ... 247,212 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {187!} _| {that is, 247,212 = the 64th root of 187!, rounded down, so insert an existing four nines solution for 187 here} ... 268,289 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {188!} _| {that is, 268,289 = the 64th root of 188!, rounded down, so insert an existing four nines solution for 188 here} ... 291,188 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {189!} _| {that is, 291,188 = the 64th root of 189!, rounded down, so insert an existing four nines solution for 189 here} ...

294,204

= |_π^11_| ... 295,259 = |_ sqrt 14! _| {but, as we see near the top, we can make 14 with four nines, so we can make 295,259 with four nines} ... 316,067 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {190!} _| {that is, 316,067 = the 64th root of 190!, rounded down, so insert an existing four nines solution for 190 here} ... 343,099 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {191!} _| {that is, 343,099 = the 64th root of 191!, rounded down, so insert an existing four nines solution for 191 here} ... 372,475 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {192!} _| {that is, 372,475 = the 64th root of 192!, rounded down, so insert an existing four nines solution for 192 here} ... 404,398 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {193!} _| {that is, 404,398 = the 64th root of 193!, rounded down, so insert an existing four nines solution for 193 here} ... 439,092 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {194!} _| {that is, 439,092 = the 64th root of 194!, rounded down, so insert an existing four nines solution for 194 here} ... 476,801 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {195!} _| {that is, 476,801 = the 64th root of 195!, rounded down, so insert an existing four nines solution for 195 here} ... 517,790 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {196!} _| {that is, 517,790 = the 64th root of 196!, rounded down, so insert an existing four nines solution for 196 here} ... 531,342 = (9^(3!)) - 99 ... 531,360 = (9-to-the-((sqrt 9)!)) - (9x9) ... 531,440 = (9-to-the-((sqrt 9)!)) - (9/9) 531,441 = (9-to-the-((sqrt 9)!)) x (9/9) 531,441 = (9-to-the-((sqrt 9)!)) + (9/9) ... 531,441 = (9-to-the-((sqrt 9)!)) + (9x9) ... 531,540 = (9-to-the-((sqrt 9)!)) + 99 ... 562,348 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {197!} _| {that is, 562,348 = the 64th root of 197!, rounded down, so insert an existing four nines solution for 197 here} ... 610,788 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {198!} _| {that is, 610,788 = the 64th root of 198!, rounded down, so insert an existing four nines solution for 198 here} ... 663,453 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {199!} _| {that is, 663,453 = the 64th root of 199!, rounded down, so insert an existing four nines solution for 199 here} ... 720,715 = |_ sqrt sqrt sqrt sqrt sqrt sqrt {200!} _| {that is, 720,715 = the 64th root of 200!, rounded down, so insert an existing four nines solution for 200 here} ...

924,269

= |_π^12_| ... 970,299 = 99-to-the-(sqrt 9) - sqrt 9 970,297 = 99-to-the-(sqrt 9) - |- sqrt sqrt 9 -| [ceiling] 970,298 = 99-to-the-(sqrt 9) - |_ sqrt sqrt 9 _| 970,299 = 99-to-the-(sqrt 9) [with just 3 nines] {99 cubed} 970,300 = 99-to-the-(sqrt 9) + |_ sqrt sqrt 9 _| 970,301 = 99-to-the-(sqrt 9) + |- sqrt sqrt 9 -| [ceiling] 970,302 = 99-to-the-(sqrt 9) + sqrt 9 ... 1,000,000 = {10 with a single π} ^ |_π_|! {with two π} ...

2,903,677

= |_π^13_| ...

9,122,171

= |_π^14_| 10,000,000 = {10 with a single π} ^ (|_π_| + |-π-|) {with three π} ...

28,658,145

= |_π^15_| = |_π^(3x5)_| ... 43,046,712 = ((9-to-the-9)/9) - 9 ... 43,046,715 = ((9-to-the-9)/9) - (sqrt 9)! ... 43,046,718 = ((9-to-the-9)/9) - (sqrt 9) ... 43,046,721 = (9-to-the-9)/9 (9-to-the-8th only uses 3 9s so far) ... 43,046,724 = ((9-to-the-9)/9) + (sqrt 9) ... 43,046,727 = ((9-to-the-9)/9) + (sqrt 9)! ... 43,046,730 = ((9-to-the-9)/9) + 9 ...

90,032,220

= |_π^16_| = |_ π^(|-π-| x |-π-|) _| 100,000,000 = {10 with a single π} ^ (|-π-| + |-π-|) {with three π} ...

282,844,563

= |_π^17_| ... 387,420,488 = (9-to-the-9) - (9/9) 387,420,489 = (9-to-the-9) x (9/9) 387,420,490 = (9-to-the-9) + (9/9) 387,420,491 = 387,420,492 = (9-to-the-9) + (9/(sqrt 9)) 387,420,493 = 387,420,494 = 387,420,495 = (9-to-the-9) + 9 - (sqrt 9) 387,420,496 = 387,420,497 = 387,420,498 = (9-to-the-9) + ((sqrt 9)x(sqrt 9)) 387,420,499 = 387,420,500 = (9-to-the-9) + 9 + (sqrt 9) 387,420,501 = 387,420,502 = 387,420,503 = (9-to-the-9) + 9 + (sqrt 9)! 387,420,504 = 387,420,505 = 387,420,506 = (9-to-the-9) + 9 + 9 ... 387,420,570 = (9-to-the-9) + (9 x 9) ... 387,420,588 = (9-to-the-9) + 99 ... 774,840,978 = (9^9) + (9^9) ... 888,582,403 = |_π^18_| = |_ π^(|_π_| x |_π_|!) _| 1,000,000,000 = {10 with a single π} ^ (|_ π x π _|) {with three π} 2,791,563,949 = |_π^19_| 8,769,956,796 = |_π^20_| 10,000,000,000 = {10 with a single π}^{10 with a single π} {with two π} 27,551,631,842 = |_π^21_| 86,556,004,191 = |_π^22_| 271,923,706,893 = |_π^23_| 854,273,519,913 = |_π^24_| 2,683,779,414,317 = |_π^25_| 8,431,341,691,876 = |_π^26_| 26,487,841,119,103 = |_π^27_| 83,214,007,069,229 = |_π^28_| 261,424,513,284,460 = |_π^29_| 821,289,330,402,748 = |_π^30_| 2,580,156,526,864,955 = |_π^31_| 8,105,800,789,910,700 = |_π^32_| 25,465,124,213,045,796 = |_π^33_| 80,001,047,150,456,241 = |_π^34_| and so... to infinity!

A Discussion of Some Deeper Mathematical Issues Related to the Four Nines Puzzle

There is, of course, no upper limit to the numbers which we can build with nines and the operators mentioned at the top of this web page. Consider the infinite series: 9 9! (9!)! ((9!)!)! (((9!)!)!)! ((((9!)!)!)!)! ... and that just uses one 9. Playing around with the puzzle, it soon becomes obvious that there are integers that can be represented by four nines in an infinite number of different ways. It is NOT obvious whether ALL numbers can be represented in at least one way. We return to this later. By more advanced mathematics, it might be shown that every integer can be represented by a sufficiently long sequence of the operators we use here.

Dudeney Invented the Four Nines; Knuth Criticizes Dudeney

In "The Weekly Dispatch" of 4 February 1900, the puzzle column by Dudeney introduced this problem. But Professor Donald Knuth comments on Dudeney's Solution Number 310, which gives a table. Knuth criticizes: "he disallows (sqrt 9)! for completely illogical reasons; also, he fails to express 38, 41, 43, ... with fewer than five 9s." Knuth on Dudeney

Definition of Factorial

Let us note that there are some unsolved mathematical questions about the Factorial Function N! The well-known definition is: 1! = 1 2! = 1 x 2 = 2 3! = 1 x 2 x 3 = 6 4! = 1 x 2 x 3 x 4 = 24 5! = 1 x 2 x 3 x 4 x 5 = 120 6! = 1 x 2 x 3 x 4 x 5 x 6 = 720 7! = 1 x 2 x 3 x 4 x 5 x 6 x 7 = 5040 8! = 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 = 40,320 9! = 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 = 362,880 10! =1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 x 10 = 3,628,800 ... and so on

Any Integer is the First Digits of Some Factorial

Believe it or not, your telephone number, your Social Security Number, or any other whole number important to you can be found as the beginning digits of some enormous factorial. John E. Maxfield proved this theorem: "If A is any positive integer having M digits, there exists a positive integer N such that the first M digits of N! constitute the integer A." ["A Note on N!", John E. Maxfield. Kansas State University, Mathematics Magazine, March-April 1970, pp.64-67; supported in part by a National Science Foundation grant].

Factorials of Negative, Fractional, and Complex Numbers

For non-integers, and for complex numbers, the universally accepted generalization of the Factorial Function is the Gamma Function. Discussion of that shall be reserved until I provide my analysis of the Four Nines Puzzle as applied to Complex Numbers, such as first appear with sqrt (-9) = 3i.

Square-less-one Factorials

In our solutions to the Four Nines Puzzle, we often add to or subtract something from a factorial, and then take a square root. Typically, this is not a whole number, so we round down, or round up (floor or ceiling). This leads to the question: When is N! + 1 a perfect square? Or, equivalently, When is sqrt (N! + 1) a whole number? The only solutions known are: 25 = 4! + 1 = 5 x 5 121 = 5! + 1 = 11 x 11 5041 = 7! + 1 = 71 x 71 Nobody knows if there are any other perfect squares one more than factorials. Clearly related is a paper by Berend and Osgood: [Journal of Number Theory, vol. 42, 1992] which proves that for any polynomial P of degree > 1 the set of positive integers N for which P(X) = N! has an integral solution X, is of zero density. This paper explicitly says that it is not known if the equation X^2 - 1 = N! has only finitely many solutions. We also don't know if there are an infinite number of primes of the form n! + 1. We also don't know if there are an infinite number of primes of the form n! - 1. The largest such prime that we know is 3610! - 1. It has 11,277 digits [Caldwell, title to be added here, 1993] By the way, there is a fairly elementary proof that, except for 0!=1 and 1!=1, NO factorials are perfect squares. That fairly elementary proof, though, uses a heavy mathematical result known as Bertrand's Postulate, also known as Chebychev's Theorem (after the man who first proved it). This Theorem says that there always exists at least one prime between N and 2N, if N>2. Erdos gave a genuinely elementary (although neither short nor obvious) proof. This will all be inserted here or linked to in a later version of this web page.

Ergodic Hypothesis

The Four Nines Puzzle itself is, to be sure, very elementary stuff. So I stand meekly in the shadow of the great mathematician's who were my Teachers' teachers' teachers... Gottfried Leibnitz, Jacob Bernoulli, Johann Bernoulli, Leonhard Euler, Joseph Louis Lagrange, Simeon Poisson, Pafnuty Lvovich Chebyshev, Andrei A. Markov, G. H. Hardy, Alonzo Church, David Hilbert, Norbert Wiener, Alan Turing... With them in mind, I remark that Donald Knuth conjectures that ALL integers can be made with a sufficiently lengthy combination of square roots and factorials and floors and ceilings ... built around a single 4. I make the related conjecture, based on the number 9. As we see near the top of this web page, we can make a 4 from a single 9, with a lot of square roots, factorials, and floor functions. Hence Knuth's conjecture for 4 immediately applies to 9. In summary of a subtle proof of Knuth's Conjecture, still in progress, factorials make a number bigger, and square roots make it smaller. Iterating sufficiently, we are "folding" the algebraic number line back onto itself recursively, and this is an ergodic property, which carries a number arbitrarily close to any given integer, at which point a final floor or ceiling gets us exactly to that given integer.

Complexity Ordering of Solutions

The problem becomes (as I shall show in a forthcoming paper co-authored by Andrew Carmichael Post and Dr. George Hockney) how to achieve as many integers as possible with 4 nines -- for a given degree of "complexity" as defined by the number of symbols of a standardized way of expressing the combination of 9, 99, 999, 9999, +, -, x, /, sqrt, factorial, floor, and ceiling (say in Backus-Naur Form). For example: 9999 = 9999 (complexity = 4) 1 = 99/99 (complexity = 5) 1008 = 999 + 9 (complexity = 5) 9801 = 99 x 99 (complexity = 5) 2 = 99/9 - 9 (complexity = 6) 20 = 99/9 + 9 (complexity = 6) 19 = 9 + 9 + 9/9 (complexity 7) 36 = 9 + 9 + 9 + 9 (complexity 7) 13 = 9 + sqrt 9 + 9/9 (complexity 8) 40 = |_ sqrt 999 _| + 9 (complexity 8) and so on. The complexity function creates an order on the solutions to the four nines problem. Of interest are such functions as the smallest number whose complexity exceeds a given value, and upper and lower bounds on the ratio of a number to its complexity. Almost all numbers have very high complexity. But details will be revealed in that forthcoming paper. The problem of whether two strings of characters evaluate to the same integer is a very hard problem, in terms of the amount of computation necessary to determine it in general, called the "word problem" in complexity theory.

The "Four Nines Problem" is closely related to the "Four Fours Problem"

The "Four Fours Problem" first appeared in: "Mathematical Recreations and Essays", by W. W. Rouse Ball [1892]. In this book the "Four Fours Problem" is called a "traditional recreation." There are several fine sites on the World Wide Web for "Four Fours Problem." But I recommend to the reader: "Mathematical Games", by Martin Gardner, [Scientific America, Jan 1964]. The [Feb 1964] issue has answers to the puzzles posed in January. Martin Gardner was extending the "two fours" problem as first posed by J. A. Tierney in 1944, and extended by others in 1964. ["E631", J. A. Tierney, Amer. Math Monthly, 52(1945)219]. ["64 Ways to Write 64 Using four 4's", M. Bicknell and V. E. Hoggatt, Recreational Mathematics Magazine, 14(1964)13-25]. More recently, we have Knuth's Conjecture: "Representing Numbers Using Only One 4", Donald Knuth, [Mathematics Magazine, Vol. 37, Nov/Dec 1964, pp.308-310]. Knuth shows how (using a computer program he wrote) all integers from 1 through 207 may be represented with only one 4, varying numbers of square roots, varying numbers of factorials, and the floor function. For example: Knuth shows how to make the number 64 using only one 4: |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt sqrt |_ sqrt |_ sqrt |_ sqrt sqrt sqrt sqrt sqrt (4!)! _| ! _| ! _| ! _| ! _| ! _| ! _| ! _| As to notation in the above example, he means sqrt n! stands for sqrt (n!), not (sqrt n)! Knuth further points out that |_ sqrt |_ X _| _| = |_ sqrt X _| so that the floor function's brackets are only needed around the entire result and before factorials are taken. He CONJECTURES that all integers may be represented that way: "It seems plausible that all positive integers possess such a representation, but this fact (if true) seems to be tied up with very deep propertis of the integers." Your Humble Webmaster believes that Knuth is right, for 9 as well as 4, and will prove that in a forthcoming paper. Knuth comments: "The referee has suggested a stronger conjecture, that a representation may be found in which all factorial operations precede all square root operations; and, moreover, if the greatest integer function [our floor function] is not used, an arbitrary positive real number can probably be approximated as closely as desired in this manner." If we abbreviate the long strings of symbols, we can express this more elegantly. Let NK be the number (...((4!)!)!...)! with K factorial operations. Then the anonymous referee's stronger conjecdture is equivalent to: log NK/(2^|_log(log NK)_|, for K=1,2,3..., are dense in the interval (1,2). Notationally here log means log to the base 2, and ^ means exponentiation. One key to the proof I shall publish is that, for any logarithm base: "The fractional part of log N is dense on the unit interval." From that, it has been proved that: "The fractional part of log N! is dense on [0,1]." ["A Note on N!", John E. Maxfield. Kansas State University, Mathematics Magazine, March-April 1970, pp.64-67; supported in part by a National Science Foundation grant]. Maxfield also proves, in the notation above, if we also define logK is the Kth iterant of Log N, i.e. log2(N) = log log N, that: The fractional part of logK (NK) is dense on the unit interval.

Square Root of a Sum of Square Roots

In our solutions of the Four Nines Puzzle, we sometimes take a square root of the sum of two slightly different square roots. As noted in Mathpages Mathpages #305: How can we find integer solutions of M = sqrt ( sqrt (N) + sqrt ((KxN)+1)) "This can be viewed as a Pell Equation with an extra solution on the solution." We have: M^2 = sqrt (N) + sqrt ((KxN)+1) in integers, so we know that there exist integers X and Y such that: N = Y^2 KN + 1 = X^2 X + Y = M^2 Eliminating N from the first two equations above gives the Pell equation: X^2 + KY^2 = 1 For any given K we are looking for solutions X,Y such that X+Y is a square. "Of course, for any positive integer K there are infinitely many solutions to the Pell Equation, but solutions with X+Y=square are rare. For example, with K=8, the values of X+Y satisfying the Pell Equation" are: ((16 + 11 sqrt 2)/8) (3 + sqrt 8)^Q + ((16 - 11 sqrt 2)/8) (3 - sqrt 8)^Q This gives the sequence 4, 23, 134, 781, 4552 ..., which satisfies the second-order recurrence: S(j) = 6S(j-1) - S(j-2) "So the question is whether this sequence contains any squares after the initial vale 4. Recall the proof that the only square Fibonacci numbers are 0, 1, and 144." [J. Cohn, "On the Square Fibonacci Numbers", J. London Math Soc., 39 (1964) 537-540] "In general the problem reduces to finding square terms of a general second-order recurring sequence, like the Fibonacci sequence. The best approach might be to apply Cohn's method of proof to the general second-order recurrence."

Square Roots of Factorials in Quantum Mechanics

In our solutions to the "Four Nines Problem" we often take square roots of factorials, and sometimes square roots of square roots of factorial of factorials, and so forth. As it turns out, there are some solved and some unsolved problems in the Physics of Quantum Wave Functions which involve square roots of factorials. As B. Nagel has written, "Mathematical problems in Quantum Optics", {ref to be done}: "As a continuation of earlier studies of squeezed states and other special harmonic oscillator states of interest in quantum optics I have studied the phase representations of these states. Although the phase observable, which is roughly speaking conjugate to the number operator, does not exist as a hermitian operator -- this is a longstanding and still popular problem, initiated by Dirac in 1927 -- it exists as a so-called general observable and permits a probability interpretation via a phase distribution on the unit circle. The corresponding wave function is obtained simply by substituting the harmonic wave exp(in[[phi]]) for the number state |n>. The series expansions thus obtained for the coherent and squeezed states contain a square root of a factorial n!, which makes it impossible to get closed analytical expressions. Approximate analytical expressions have been derived, valid for large values of the mean value of the number operator...." For another example, see the computer program given in: Grozin Hydrogen Wave Functions and E1 Transitions Procedure R(n,1); % radial wave function -2/n^2*sqrt(factorial(n+1)/factorial(n-1-1))*exp(-r/n) ... etc. Wouldn't it be interesting if an in-depth analysis of the century-old recreation math puzzle about Four Nines turned out to be useful in solving a problem in 21st Century Quantum Optics with Squeezed States? {more discussion to be added in February 2004} Special Thanks to Dr. George Hockney, NASA/JPL, for informal discussion and review in January-February 2004. Thanks to Forrest Bishop for informal discussion and review in January-February 2004.

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