Problem A. 486. (September 2009)
A. 486. Denote by (n) the exponent of 2 in the prime factorization of n!. Show that for arbitrary positive integers a and m there exists an integer n>1 for which .
(5 pont)
Deadline expired on 12 October 2009.
Solution. We will use the wellknown fact that .
If the base2 form of the number n is , where the digits are all 0 or 1, then
Now we will show that there exists an integer r which is relatively prime to m, and an infinite sequence of positive integers such that
Let m=2^{t}u where u is odd, and consider an arbitrary positive integer it for which (u) divides i1. By the EulerFermat theorem,
 (1) 
and, due to it,
 (2) 
The relations (1) and (2) determine the residue class of 2^{i}1 modulo m, and it must be relatively prime to m.
Since m and r are relatively prime, there is a positiv integer u such that . For we achieve
Based on the solution of András Éles
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