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Old 2015-10-03, 20:35   #1
MattcAnderson
 
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"Matthew Anderson"
Dec 2010
Oregon, USA

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Wink 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
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Old 2015-10-03, 20:51   #2
science_man_88
 
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"Forget I exist"
Jul 2009
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Quote:
Originally Posted by MattcAnderson View Post
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
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Old 2015-10-03, 21:00   #3
R. Gerbicz
 
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"Robert Gerbicz"
Oct 2005
Hungary

2·3·11·23 Posts
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Quote:
Originally Posted by MattcAnderson View Post
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.
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Old 2015-10-04, 02:33   #4
LaurV
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Ye spoiled all the fun...
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