20151003, 20:35  #1 
"Matthew Anderson"
Dec 2010
Oregon, USA
1009_{10} 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 
20151003, 20:51  #2  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
Quote:
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 20151003 at 20:52 

20151003, 21:00  #3  
"Robert Gerbicz"
Oct 2005
Hungary
1,531 Posts 
Quote:


20151004, 02:33  #4 
Romulan Interpreter
"name field"
Jun 2011
Thailand
2·11·449 Posts 
Ye spoiled all the fun...

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