mersenneforum.org exponential growth patterns
 User Name Remember Me? Password
 Register FAQ Search Today's Posts Mark Forums Read

 2015-10-03, 20:35 #1 MattcAnderson     "Matthew Anderson" Dec 2010 Oregon, USA 4A816 Posts exponential growth patterns Hi Math People, Are there integers n and m such that 2^m and 2^n have the same number of digits and the sum of the digits for 2^n and 2^m is the same? Maple code > a := Vector(30); > for b to 30 do a[b] := 2^b end do; > > print(n, 2^n, "numdigits in", 2^n, digitsum*of*2^n); for c to 30 do print(c, a[c], floor(log10(a[c]))+1, add(k, k = convert(a[c], base, 10))) end do; Regards, Matt
2015-10-03, 20:51   #2
science_man_88

"Forget I exist"
Jul 2009
Dartmouth NS

2×3×23×61 Posts

Quote:
 Originally Posted by MattcAnderson Hi Math People, Are there integers n and m such that 2^m and 2^n have the same number of digits and the sum of the digits for 2^n and 2^m is the same? Maple code > a := Vector(30); > for b to 30 do a[b] := 2^b end do; > > print(n, 2^n, "numdigits in", 2^n, digitsum*of*2^n); for c to 30 do print(c, a[c], floor(log10(a[c]))+1, add(k, k = convert(a[c], base, 10))) end do; Regards, Matt
well if 2^n and 2^m are the same number of digits n and m can't differ by more than 3 at last check since 2^4 >10
we then have the cases:

m=n+1 \\ 2^m=2*2^n
m=n+2 \\ 2^m= 4*2^n
m=n+3 \\ 2^m= 8*2^n

so it then comes down to how many digits affect others because:
1) 1 won't go past ten under anything but a carry from another place.
2) 2, won't go past ten except in that 3rd case or a carry.
3) 3 and 4 only break past ten in all but the first case of the three above or a carry
4) the rest go past ten regardless and 5 only affects 1 when it's the second case or higher, etc.

Last fiddled with by science_man_88 on 2015-10-03 at 20:52

2015-10-03, 21:00   #3
R. Gerbicz

"Robert Gerbicz"
Oct 2005
Hungary

5·17·19 Posts

Quote:
 Originally Posted by MattcAnderson Hi Math People, Are there integers n and m such that 2^m and 2^n have the same number of digits and the sum of the digits for 2^n and 2^m is the same? Maple code > a := Vector(30); > for b to 30 do a[b] := 2^b end do; > > print(n, 2^n, "numdigits in", 2^n, digitsum*of*2^n); for c to 30 do print(c, a[c], floor(log10(a[c]))+1, add(k, k = convert(a[c], base, 10))) end do; Regards, Matt
There is only trivial solution (n=m). Let s(k) the sum of the digits of k, then s(k)-k is divisible by 9, so if s(2^n)=s(2^m) then 9 divides 2^m-2^n, say n<m there are only at most 3 possible cases (if 2^n and 2^m has the same number of digits): m=n+1,n+2,n+3. For these 2^m-2^n={1,3,7}*2^n so not divisible by 9.

 2015-10-04, 02:33 #4 LaurV Romulan Interpreter     "name field" Jun 2011 Thailand 3×23×149 Posts Ye spoiled all the fun...

 Similar Threads Thread Thread Starter Forum Replies Last Post CuriousKit Miscellaneous Math 24 2015-04-06 18:40 Xyzzy mersennewiki 1 2010-12-01 23:05 davieddy Puzzles 10 2010-05-25 03:43 wustvn Puzzles 7 2008-11-20 14:00 SK8ER-91823 Twin Prime Search 4 2007-04-14 12:52

All times are UTC. The time now is 04:47.

Wed Feb 8 04:47:39 UTC 2023 up 174 days, 2:16, 1 user, load averages: 0.63, 0.74, 0.75

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