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Friends, we have a towering puzzle in which integers are equated to ordered pairs of colors. The ordered pairs are based on the nine colors given below:

White, yellow, blue, red, violet, orange, green, maroon, black

Hints:

  1. The numbers are given in base ten, but the color coding does not correspond to base ten digits.

  2. The nature of the numbers involved in the calculation and the ordering of the list of colors are suggested in the title.

enter image description here

Spoiler #1:

From the Broadway musical The Music Man: "Oh we've got trouble/Right here in River City/With a capital T that rhymes with P/And that stands for pool!" So the "game" is billiards, and the nine colors correspond to the cue ball (taken as 0) and balls numbered 1 through 8 in order.

Spoiler #2:

With only nine colors, they must correspond to digits in base nine rather than ten.

Spoiler #3:

The towers are infinitely high, but they give well-defined results for all positive whole number bases as well as $-1$.

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1 Answer 1

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Edit: I've included both the final answer here and my original attempt as I think it's interesting that both line up for the examples given (you can see how both answers are linked but not quite the same).

First step (revealed in first spoiler)

Each colour corresponds to a digit with the correspondence indicated by the value of solid colours in billiards: yellow=1, blue=2, red=3, violet=4, orange=5, green=6, maroon=7, black=8 and we assign white (cue ball) to 0.

Second step (revealed in second spoiler)

Each ordered pair of colours corresponds to a two digit number in base 9 (since we only have nine colours), $2$ maps to $37$ in base $9$ which is $34$ in base $10$, etc.

Third step

The title indicates the word "Tower" with a capital T that rhymes with P, aka Power. What we are really looking at are "power towers" - that is sequences of the form $b^{b^{b^{b^\ldots}}}$. In particular, it turns out that for any integers $b$ and $k$, the power tower of $b$ converges modulo $k$ so it makes sense to talk about the infinite power tower modulo $k$.

Overall

For each integer $K$, we need to find the value to which the power tower converges, modulo $81$. Then we convert that number to base $9$ and rewrite it as an ordered pair of colours.

The OP has shown me a slick way to work this out which involves first working out the values of the iterated Carmichael function of $81$ and then, proceeding backwards, working out the iterated power tower modulo each of these integers in turn.
Firstly, $\lambda(81) = 54\,\,,\,\, \lambda(54) = 18\,\,,\,\, \lambda(18) = 6\,\,,\,\, \lambda(6) = 2 $

On that basis, I believe the solutions for the missing values are as follows (I'm not totally confident on this process so if I have done something wrong, please correct me).

$5 \equiv 1$ mod $2$
$5^1 \equiv 5$ mod $6$
$5^5 \equiv 11$ mod $18$
$5^{11} \equiv 29$ mod $54$
$5^{29} \equiv 56$ mod $81$
This implies that the iterated power tower for $5$ converges to $56$ modulo $81$ which is $62$ in base $9$, represented by green blue.

$10 \equiv 2$ mod $2$
$10^2 \equiv 4$ mod $6$
$10^4 \equiv 10$ mod $18$
$10^{10} \equiv 10$ mod $54$
$10^{10} \equiv 10$ mod $81$
The iterated power tower for $10$ converges to $10$ modulo $81$ which is $11$ in base $9$, represented by yellow yellow

$20 \equiv 2$ mod $2$
$20^2 \equiv 4$ mod $6$
$20^4 \equiv 16$ mod $18$
$20^{16} \equiv 16$ mod $54$
$20^{16} \equiv 43$ mod $81$
The iterated power tower for $20$ converges to $43$ modulo $81$ which is $47$ in base $9$, represented by violet maroon

$323 \equiv 1$ mod $2$
$323^1 \equiv 5$ mod $6$
$323^5 \equiv 17$ mod $18$
$323^{17} \equiv 53$ mod $54$
$323^{53} \equiv 80$ mod $81$
The iterated power tower for $323$ converges to $80$ modulo $81$ which is $88$ in base $9$, represented by black black

Original Answer Third step

The title indicates the word "Tower" with a capital T that rhymes with P, aka Power. This indicates that we should be looking at powers of numbers in determining the correspondence between integers.

Indeed, notice that $2^{34} \equiv 34$ (mod $81$).
Also, $4^{40} \equiv 40$ (mod $81$) and $40$ in base $10$ becomes $44$ (violet violet) in base $9$.
Trivially $1^1 \equiv 1$ (mod $81$).

Overall

For each integer $K$, we are looking for the smallest integer $n$ which solves $K^n \equiv n$ (mod $81$), then we need to convert $n$ to base $9$ and rewrite it as an ordered pair of colours.

There are some subtleties here.
For example, $3^{81} \equiv 81$ (mod $81$) but to get this solution, we have to go to $81$ which is the same as $00$ (white white) (mod $81$).
Furthermore $(-1)^{-1} \equiv -1$ (mod $81$) so in this case we don't go to the next solution of $82$. Of course, $-1$ here is represented the same way as $80$ which is $88$ (black black) in base $9$.
Finally, $0$ is undefined because it maps to no pair of colours.

On that basis, here are the solutions

$5^{58} \equiv 58$ (mod $81$), $58$ is $64$ in base $9$ which is green violet.
$10^{10} \equiv 10$ (mod $81$), $10$ is $11$ in base $9$ which is yellow yellow.
$20^{41} \equiv 41$ (mod $81$), $20$ is $22$ in base $9$ which is blue blue.
$323$ is slightly unusual as $323^{82} \equiv 82$ (mod $81$), which is represented by white yellow. I don't consider $-1$ here as a valid solution because I'm considering powers used in the normal sense and not in the modulo arithmetic sense (I think this would open up a can of worms otherwise).

Connection

It seems like the two answers should line up because the solution in the original would seem to line up with being the limit point. This is why we get some of the answers to correspond but this won't work in general essentially because
$n \equiv m$ mod $k$
does not imply
$K^n \equiv K^m$ mod $k$
as pointed out by the OP

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  • $\begingroup$ For $-1$ I consider is quite legit to use the complements form, which is obvious because the power tower converges without using modulo arithmetic or p-adics for that case (also for +1). Thus $...88$ = (black, black). $\endgroup$ Commented May 16 at 15:04
  • $\begingroup$ @OscarLanzi I'm not sure I fully understand the iterative process, can you show how it works for $2$? $\endgroup$
    – hexomino
    Commented May 16 at 15:57
  • $\begingroup$ $2^2\equiv4\bmod 6$ (you need even residues so $2^2$ instead of $2^0$), $2^4\equiv16\bmod18,2^{16}\equiv34\bmod54,2^{34}\equiv34\bmod81$, so the power tower $2^{2^{2^2...}}$ will converge to $34\bmod81\implies37$ base nine. $\endgroup$ Commented May 16 at 16:05
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    $\begingroup$ @OscarLanzi I've updated this now, sorry for not getting back to it previously. $\endgroup$
    – hexomino
    Commented May 23 at 15:04
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    $\begingroup$ The iterated Carmicharl function always ends with $2$, so start with terms being all even or all odd. Then go with $n^2$ or $n\bmod 6$ accordingly since in this case the Carmichael functions are $81\to54\to18\to6\to2$. $\endgroup$ Commented May 23 at 16:27

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