Thursday, 31 October 2013

number theory - A positive integer is equal to the sum of digits of a multiple of itself.



Let n be a positive integer, prove there is a positive integer k so that n is equal to the sum of digits of nk.




I'm not really sure how I should approach this problem, I tried to do a constructive approach but I got lost.



I tried just proving existence but that didn't work either.



I'm sorry if this doesn't look like a put work into it but I feel like nothing I have done is going to yield any results. So I hope you guys can solve this problem.



Thank you very much in advance, regards.


Answer



There is a simple way to answer this sort of question. And there is a related type of number called Niven numbers.




A Niven number is a number, N, for which its sum of digits divides N. What you are trying to prove is that every integer is the sum of digits of a Niven number.



Here is how we can go about it. Every number can be written in the form: N=bk10k+bk110k1++b0100 for some k and 0bi9.



Since 10i can take at most n values modulo n, we have 10i+j=10i for some j and all i. In other words the function f(i)=10i has period j modulo n.



Thus if we consider N=10nj+10(n1)j++10j we have N \equiv 10^{nj} + 10^{(n-1)j} + \cdots + 10^{j} \equiv (\underbrace{1+1+\cdots+1}_{n \text{ times}})10^{j} \equiv n10^{j} \equiv 0 \mod n.



Ivan Niven introduced Niven numbers as a possible avenue of research accessible to undergraduates. His paper appeared in a journal for undergraduate education. These numbers also go under the name of Harshad numbers, but I believe Niven numbers are more common. The Wiki page for "Harshad" numbers contains many references to papers concerning Niven numbers.



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