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Topic: Digit sum (Read 2719 times) |
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Christine
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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.
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towr
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Re: Digit sum
« Reply #1 on: Feb 4th, 2013, 10:37pm » |
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So, if I understand correctly we're checking for dsum(np)p = n, or replacing n by kp, dsum(kp^2) = k
The value of kp^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)
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| « Last Edit: Feb 4th, 2013, 10:41pm by towr » |
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Christine
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Re: Digit sum
« Reply #2 on: Feb 7th, 2013, 9:41am » |
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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?
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| « Last Edit: Feb 7th, 2013, 9:49am by Christine » |
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towr
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Re: Digit sum
« Reply #3 on: Feb 7th, 2013, 11:11am » |
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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.)
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Christine
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Re: Digit sum
« Reply #4 on: Feb 7th, 2013, 2:40pm » |
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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?
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Francesc
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Re: Digit sum
« Reply #5 on: Mar 7th, 2013, 7:23am » |
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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
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| « Last Edit: Mar 7th, 2013, 7:24am by Francesc » |
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