2022-11-07, 21:51 | #12 |
Dec 2008
you know...around...
929 Posts |
Take it, it's free
While this thread is still on the front page, let me drop this number here:
Code:
977501592599998899778899999999987869899997877899798979989989676889999889999899775899989798999998789769998994897988599999999988989899879989699897999798898997599889876898998999994977897978999988999939998887 |
2023-02-27, 20:19 | #13 |
Dec 2008
you know...around...
929 Posts |
Fifth powers riddle
This could make for a nice puzzle, or at least serve as inspiration for further endeavors:
(assume all mentioned variables being positive integers) For any base b, are there only finitely many numbers x not divisible by b such that the sum of digits of x is larger than the sum of digits of \(x^5\) (in base b)? Or is there a threshold \(b_0\) above which all \(b > b_0\) can have infinitely many (or at least one) such solutions? (Is \(b_0\)=283 for fifth powers?) Some solutions for \(b \leq 100\) (searched up to \(x=10^8\)): Code:
b x 8 4* 27 9*, 23 32 2*, 4*, 8*, 16* 39 177716 40 20* 53 8210 54 18*, 36*, 138* 55 31 60 42 64 16*, 32*, 48*, 245408* 72 36* 77 822816 79 16255431 90 299047 92 52881676 96 24*, 48*, 6600*, 17256* 98 7140* * semi-trivial solutions, since b|x^5 b=27 seems to yield the smallest non-trivial solution. Or might there be a smaller b for which such a solution can be found? What about higher prime powers p? It is trivial to show that there are arbitrarily large bases b for which \(x \geq 3\) (mutually, all \(x^n\) for \(1 \leq n \leq p-1\)) is a solution whenever \(b=x^p+2-x\). Excluding all those trivial and semi-trivial solutions, for p=7 so far the smallest (in terms of b) non-trivial solution I found was b=492, x=121820. Is it possible to find a non-trivial solution for larger p? Have I overlooked a way to trivially construct solutions? |
2023-05-28, 18:21 | #14 |
Dec 2008
you know...around...
929 Posts |
Useless as usual
As of today, 10k views and no more replies.
Um... go me? |
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