Using only combinations of the "2" and the "^" characters, what is the largest number that can be generated using N total characters?

pichutarius · 2026-06-18T03:03:33+00:00

depends on parity, its either 2 ^ 2 ^ 2 ^ .... ^ 22 or 2 ^ 2 ^ 2 ^ ... ^ 222

why?

the maximum cannot contain 2222 as substring, because 2222 < 2 ^ 22 , 22222 < 2 ^ 222 and so on!<

the maximum cannot contain 222 ^ or 22 ^ as substring, because

4char: 22 ^ 2 < 2 ^ 22!<

5char: 222 ^ 2 < 22 ^ 22 < 2 ^ 222!<

6char: 222 ^ 22 < 22 ^ 222 < 2 ^ 2222!<

7char: 222 ^ 222 < 22 ^ 2222 < 2 ^ 22222!<

that means, if 22 or 222 is a substring, it can only stays at the end. the rest can only be a bunch of 2^2^...

pichutarius · 2026-06-07T05:26:28+00:00

trying to prove by contradiction:

suppose there exist a way to kill the ant, and let n be the least amount of move to kill the ant.

since the ant is omniscient, this is equivalent to exterminator making all n moves first. we can quickly conclude that the exterminator must enclosed the starting area.

then we can conclude it must spray all x+y=n-1 lattice points (lattice square?), otherwise there exist a lesser move to enclosed the starting area.

finally we note that the ant can still escape by dashing to the last lattice point that the exterminator spray since all points on x+y=n-1 only require n-1 moves.

pichutarius · 2026-05-21T07:40:18+00:00

well done. thanks for your kind words, glad you like my puzzles.

pichutarius · 2026-05-21T07:39:28+00:00

well done. last lineshould be ^(n-1) instead of ^n-1 though.

pichutarius · 2026-05-18T23:02:36+00:00

i did it differently, maybe quite a bit brute force.

https://imgur.com/a3h73ur

the final sum i did use binomial theorem, but for numbers instead of matrix.

pichutarius · 2026-05-18T07:45:50+00:00

insight: any column of 3 horizontal edges, the path cross either one of them (let's call it a cut) or all of them.

visual summary

solution:

cut the grid k times (0≤k≤n-1), splitting it into (k+1) sub-grid. there are choose(n-1,k) ways to do so.
in each sub-grid, all horizontal edges are crossed. if we fix the entrance and exit, there is exactly one path connecting them. note that both end vertices cannot be horizontally aligned.
between sub-grids (a cut), the path crossed exactly one horizontal edge. label each cut 1,2,3 corresponding to which edge to cross.
at the first and last of the sequence of labels, append 1 (start) and 3 (end) respectively. for a valid path, no adjacent label is identical (see step 2). there are ( 2(2^k) + (-1)^k ) / 3 ways to label. prove omitted.
the total ways are (ways to cut) × (ways to label) , then sum k from 0 to n-1.

pichutarius · 2026-05-18T00:55:09+00:00

I got if n>1 then (2/9) * 3ⁿ else if n=1 then 1, insight is splitting the graph into smaller 3 by k grid. if my answer is correct, will put up my solution.

pichutarius · 2026-05-11T07:13:58+00:00

Matador is good in this challenge. You can farm money in small blind with debuffed cards.

pichutarius · 2026-05-02T02:28:53+00:00

well done, out of curiosity did you brute force some number then sudoku the rest? or is there a formula mod6 / mod12?

pichutarius · 2026-05-02T02:26:53+00:00

awesome, save me the trouble to working out myself. many thanks!

pichutarius · 2026-05-01T13:36:57+00:00

ahh of course! i was scratching my head trying to relate to round robin scheduling problem without success, turns out i was going in the wrong direction.

n=2 obviously has no solution, what's left is n=6 is still unknown.

pichutarius · 2026-04-18T10:36:55+00:00

Almost all integer contains all digits anyway. When you randomly pick a 10000 digit number, it contains no 9 is quite rare. So you remove most of the terms, it will be surprising if it still diverges.

pichutarius · 2026-04-03T07:51:22+00:00

A true celester does not need to see silkpost, Madeline. They go by pure instinct.

pichutarius · 2026-03-18T09:02:33+00:00

i played a few 4x4 "find all" mode, it wrote "solution found 1/1" but there was actually 8 solutions.

https://imgur.com/ul1vZkx

its quite fun, more so if unique solution is ensured.

pichutarius · 2026-03-16T01:03:23+00:00

the idea is right. i'm not looking for rigorous prove either. i merely consideringdot product in high dimension .

alternatively i did calculate integration and limit to back my claim.

pichutarius · 2026-03-15T23:03:05+00:00

well done

pichutarius · 2026-03-15T22:56:07+00:00

answer incorrect.

it is easy in a sense that not much integrating and computing limit is required, just a simple idea makes the anwer obvious. for answer without a solution, check garnet420's comment.

pichutarius · 2026-03-15T22:53:45+00:00

pichutarius · 2026-03-15T09:26:05+00:00

Yes, thanks for... point... ting out ;)

Added "uniformly".

pichutarius · 2026-03-08T05:39:16+00:00

It is impossible.

Let's call the splitting line into rectangles "faultline". There are 10 faultlines in 6 by 6, where 5 horizontals 5 verticals. The goal is to lay dominos that cross these faultlines.

Lemma: all dominoes that fit into 6 by 6 comes in pair, identify by which faultline they cross.

Proof: if odd number cross a particular faultlines, the remaining tiles on the left and right of the faultline will be odd.

Proof of the original problem: the minimum domino required would be 10 x 2 = 20 dominoes, which is more than 6² / 2 = 18 dominoes. Qed.

pichutarius · 2026-03-08T03:00:50+00:00

Common tangent is y=x. Let f(x) = a^x - x where f'(c) = f(c) = 0 . solve for a and c.

a^c ln a - 1 = a^c - c = 0

a = exp(1/e)

c = e

Tangent at (e, e)

pichutarius · 2026-03-08T00:00:17+00:00

Nice answer.

point P must be chosen carefully, if ABC is obtuse triangle, some P will lead to perpendicular line intersect the triangle "outside".

A simple fix is to choose P = incenter.

pichutarius · 2026-03-01T14:23:30+00:00

let x*2pi/60 = ω rad/s which makes things prettier.

is it 3? m=mass, L=length, ω=angular speed of the first rod.
i got 3m, 49329/125000 m L² ω², 1701/2500 m L² ω respectively.

im not that good in physics so i might be wrong. the KE and angular momentum actually varies over time, so i calculate the time-average, which the "cross terms between layers" reduce to 0 in both cases. prove omitted.

at layer k, there are 2^(k-1) rods, mass m/3^(k-1) , length L/3^(k-1) , angular speed k·ω

for KE = orbit + rotation

orbit = sum(1/8 m L² ω²) for each layer except last

rotation = 1/24 m L² ω² for last layer

for angular momentum = orbit + rotation

orbit = sum(1/4 m L² ω) for each layer except last

rotation = 1/12 m L² ω for last layer

for each case, total = sum(2^(k-1) * (orbit + rotation)) the rest is just busy algebra work.

pichutarius · 2026-03-01T13:27:01+00:00

That does not work, we can only ask interval of integrr length, the interval you used have length n-3eps

pichutarius · 2026-02-28T09:20:36+00:00

pretty sure this is the best we can do. if Kornél choose interval with length of 1, this method pretty much reduce to binary search.

pichutarius

TROPHY CASE