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        riddles >> medium >> Digit sum
        
(Message started by: Christine on Feb 4^th, 2013, 11:00am)

Title: Digit sum
Post by Christine on Feb 4^th, 2013, 11:00am

How to prove or disprove:

For any integer p>0, there are only finitely many integers n>0 for which the p-th power of the digits sum of n^p is n.

Title: Re: Digit sum
Post by towr on Feb 4^th, 2013, 10:37pm

So, if I understand correctly we're checking for dsum(n^p)^p = n, or replacing n by k^p, dsum(k^{p^2}) = k
[hide]The value of k^{p^2} will always lie below 9p²*(log(k)+1), which grows only sublinearly with k; so as k grows, eventually k will be greater, and therefor there will always be a limited number of values for which the equality holds.
It doesn't even matter if you use decimal digit sum, or some other base (> 1)[/hide]

Title: Re: Digit sum
Post by Christine on Feb 7^th, 2013, 9:41am

Thanks.

Take any integer x, to what power should we raise it so that DigitSum (x^n) = y

e.g. for what value of n we have DigitSum (29^n) = 31

is there an algorithm to find out directly without trying several values?

Title: Re: Digit sum
Post by towr on Feb 7^th, 2013, 11:11am

I don't see why there would necessarily be such an n. For this specific case, if there is, n would have to be at least > 10000 (I've had the computer search for a solution). And seeing as DigitSum(29^n) is a mostly increasing function, I wouldn't expect 31 to crop up. (Can't really exclude it, because you can always have mostly zeroes in very large numbers. But it's improbable.)

Title: Re: Digit sum
Post by Christine on Feb 7^th, 2013, 2:40pm

I was thinking about twin primes

(5,7), (17,19)

5^2 = 25 ---> 2+5=7
17^2 = 289 ---> 2+8+9 = 19

So I was wondering about other primes, lesser primes related to greater primes of twin primes. Can we find other examples?

Title: Re: Digit sum
Post by Francesc on Mar 7^th, 2013, 7:23am

Ok, I can reduce the problem a little bit. digsum(x) and x have the same mod9, for every x (easy proof trying to get digsum(x) with the floor function)

So... suppose that your first prime is 9*k+t.
1.- t is not in (0,3,6) (duh, it wouldn't be prime)
2.- t is not in (1,4,7) or the second number (x+2) would be multiple of 3
3.- t=2. Then x^n=(9k+2)=9r + 2^n and y=digsum(x^n) =9r'+2^n that must be prime and, at the same time, y=x+2 => y=9r'+4 => 2^n=9s+4 => n has to be of the form 2+6p, for p natural
4.- t=5. y=9r'+5^n = 9r''+7 => n has to be 2+6p, again
5.- t=8. y=9r'+(-1)^n=9r''+1 => n has to be even.

So, the first prime is 9*k+s, with s in (2,5,8 ) and in the first two cases you need n=6p+2; while in the last one any even n could do the trick