mersenneforum.org Sum of digits of x and powers of x
 User Name Remember Me? Password
 Register FAQ Search Today's Posts Mark Forums Read

 2022-11-07, 21:51 #12 mart_r     Dec 2008 you know...around... 36816 Posts Take it, it's free While this thread is still on the front page, let me drop this number here: Code: 977501592599998899778899999999987869899997877899798979989989676889999889999899775899989798999998789769998994897988599999999988989899879989699897999798898997599889876898998999994977897978999988999939998887 It's a 204-digit prime number whose digital sum (1651) is larger than the digital sum of its cube (1639). I found it less than a day after my last post in this thread.
 2023-02-27, 20:19 #13 mart_r     Dec 2008 you know...around... 23×109 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 For $$b > 100$$, more and more non-trivial solutions appear (the next one is b=102, x=4767). 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?

 Similar Threads Thread Thread Starter Forum Replies Last Post Boltzmann brain Miscellaneous Math 2 2020-02-09 18:35 Uncwilly Lounge 15 2010-03-31 07:13 plandon Math 7 2009-06-30 21:29 nibble4bits Math 31 2007-12-11 12:56 Numbers Puzzles 3 2005-07-13 04:42

All times are UTC. The time now is 21:06.

Tue Mar 21 21:06:40 UTC 2023 up 215 days, 18:35, 0 users, load averages: 0.71, 0.79, 0.80

Copyright ©2000 - 2023, Jelsoft Enterprises Ltd.

This forum has received and complied with 0 (zero) government requests for information.

Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation.
A copy of the license is included in the FAQ.

≠ ± ∓ ÷ × · − √ ‰ ⊗ ⊕ ⊖ ⊘ ⊙ ≤ ≥ ≦ ≧ ≨ ≩ ≺ ≻ ≼ ≽ ⊏ ⊐ ⊑ ⊒ ² ³ °
∠ ∟ ° ≅ ~ ‖ ⟂ ⫛
≡ ≜ ≈ ∝ ∞ ≪ ≫ ⌊⌋ ⌈⌉ ∘ ∏ ∐ ∑ ∧ ∨ ∩ ∪ ⨀ ⊕ ⊗ 𝖕 𝖖 𝖗 ⊲ ⊳
∅ ∖ ∁ ↦ ↣ ∩ ∪ ⊆ ⊂ ⊄ ⊊ ⊇ ⊃ ⊅ ⊋ ⊖ ∈ ∉ ∋ ∌ ℕ ℤ ℚ ℝ ℂ ℵ ℶ ℷ ℸ 𝓟
¬ ∨ ∧ ⊕ → ← ⇒ ⇐ ⇔ ∀ ∃ ∄ ∴ ∵ ⊤ ⊥ ⊢ ⊨ ⫤ ⊣ … ⋯ ⋮ ⋰ ⋱
∫ ∬ ∭ ∮ ∯ ∰ ∇ ∆ δ ∂ ℱ ℒ ℓ
𝛢𝛼 𝛣𝛽 𝛤𝛾 𝛥𝛿 𝛦𝜀𝜖 𝛧𝜁 𝛨𝜂 𝛩𝜃𝜗 𝛪𝜄 𝛫𝜅 𝛬𝜆 𝛭𝜇 𝛮𝜈 𝛯𝜉 𝛰𝜊 𝛱𝜋 𝛲𝜌 𝛴𝜎𝜍 𝛵𝜏 𝛶𝜐 𝛷𝜙𝜑 𝛸𝜒 𝛹𝜓 𝛺𝜔