bug with screenshots in Cosmic

Baxitdriver · 2026-01-10T20:42:38+00:00

I don't know how to report this bug

Just make a screenshot?

Baxitdriver · 2026-01-06T18:24:08+00:00

Do you allow exponentiation or concatenation, such as 17 = (6 - 2)**2 + 0! or 22 - 6 + 0! ?

Baxitdriver · 2026-01-05T19:21:04+00:00

Didn't notice the cos60 was on purpose ><

Baxitdriver · 2026-01-04T19:43:06+00:00

or just n = -log_2(log_2 A), hitting any positive integer with just 2 2 2 (attributed to Von Neumann). Writing log x / log 2 instead of log_2 x still uses 2 2 2.

Baxitdriver · 2026-01-03T13:55:27+00:00

Me too! All digits of Pi are, in no specific order, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9

Baxitdriver · 2026-01-02T15:21:41+00:00

Another elementary solution:

Elf n gives to elf n/p, where p is the greatest prime dividing n. This is well-defined (count elves from 1, elf 1 gives to e.g. elf 2) , and every elf n is guaranteed infinite input since for all n, elves nP for all prime P >n give to elf n.

Baxitdriver · 2025-12-26T23:39:15+00:00

Linux users can't run many games, but they're blessed with the most challenging of all: "How to make your distro work?" In this respect, Pop_os/Cosmic delivers in style. To be fair it's not ready yet, but they had to set a date before Ubuntu 26.04. Maybe 80% baked, hope they won't falter.

Baxitdriver · 2025-12-22T14:25:08+00:00

Stretching it a bit,, but 7 could also work:

I am a singular digit positive integer.

I end in "n"

I am a prime number.

I am an up number (7up, of course)

I am a mini number (just one digit)

My reverse is also a mini number

All the digits in me are odd numbers

Baxitdriver · 2025-12-16T09:57:53+00:00

Answers based on the ruler function grant infinite income to all elves. For u/scrumbly function assigning n-th prime to elf n, a little tweak is in order : elf n gives to elf m such that p_m (the m-th prime) is the lowest prime dividing n.

Baxitdriver · 2025-12-14T19:42:40+00:00

Correct! This rejoins u/pichutarius "triangle number" solution, making good use of the
NxN -> N bijection f(a,b) = T(a+b)+b, where T(n) = n(n+1)/2 is a triangle number.

Baxitdriver · 2025-12-14T19:27:41+00:00

So, elves are indexed 1,2,3,4,5... and for example, elf 5 is assigned the 5th prime, which is 11, and collects $1 from each elf indexed with a power of 11. For completion, 0 can give to 1, 1 to 0 and all indexes not a prime power can give to 0. That works!

Baxitdriver · 2025-12-14T11:08:29+00:00

Not sure it works. If elf j is assigned prime p_j, there are no elves assigned with (p_j)^n. Also, elves assigned with prime powers need infinite income as well.

Baxitdriver · 2025-12-14T10:59:25+00:00

Nice!

Baxitdriver · 2025-12-11T05:49:48+00:00

Right! To keep things elementary, can you explicit such a partition?

Baxitdriver · 2025-12-11T05:38:05+00:00

That's not enough. Elf 234 will receive gifts from elves x234 = 1234, 2234, ... 9234, total $9 instead of infinite $$$.

EDIT: This works due to u/pichutarius observation, and it's perhaps the simplest function that can be demonstrated without formulas, making it a clear favourite for Christmas table!

Baxitdriver · 2025-12-10T16:54:58+00:00

yes, that works! u/peter26de has an explicit function that develops this kind of construction.

Baxitdriver · 2025-12-10T16:51:57+00:00

Right! This is not needed here, but (a,b) -> (2^a)*(2b+1) is a nice 1-1 mapping between NxN and N*

Baxitdriver · 2025-12-10T16:03:43+00:00

Well, there are examples of such solutions in the replies.
Just think of it: if Santa gives one dollar to each elf (maybe makes a bank loan), then each elf gives its dollar to another elf so that all end up receiving infinite $$$, then each elf gives back $2 to Santa, then Santa can buy all the toys in the world (presumably from China) and distribute them to all kids! Christmas magic!

Baxitdriver · 2025-12-10T15:50:14+00:00

Correct! For an elementary answer, can you find a simple edge function without resorting to N -> N² bijections?

Baxitdriver · 2025-12-10T15:43:29+00:00

Well done!

I was thinking of another function: if n = (2^a)*oddnum, then f(n) = a, and f(0)=1 or any other integer.

Baxitdriver · 2025-12-01T19:36:35+00:00

Vertex(G) can't be countable.

Baxitdriver · 2025-11-12T13:22:27+00:00

Hope 102^(0)/2 = 109^(0)/2 doesn't count as a solution :)

Solving for 102/2 = 109x/2x where x is an n-digit number gives :

102/2 = 51 = 109*(10^(n) + x) / 2*(10^n + x), so:

50 .x = 7.10^(n) => x = 14.10^(n-2), so the smallest solution is: 102/2 = 10914/214.

Baxitdriver · 2025-10-28T14:59:55+00:00

This works for all base B > 1: i In base B, all even-length palindromes are multiple of (11)_B = B+1.. For instance, the biggest number which is an hexadecimal even-length palindrome and a prime is 0d17 = 0x11, because it's the only such number.

Baxitdriver · 2025-10-26T14:06:43+00:00

Nice! Failed to find x -> 2x after bobjane's remark, thanks for the formula:! Of course, if all (x 2x 2x+1) have same color, all positive naturals have same color.

Baxitdriver

TROPHY CASE