 mersenneforum.org exponential growth patterns
 Register FAQ Search Today's Posts Mark Forums Read 2015-10-03, 20:35 #1 MattcAnderson   "Matthew Anderson" Dec 2010 Oregon, USA 5×233 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
Dumbassville

2×5×839 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

112×13 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 271D16 Posts Ye spoiled all the fun...    Thread Tools Show Printable Version Email this Page 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

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