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A. 486. Denote by \nu(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 \nu(n) \equiv a \pmod{m}.

(5 points)

Deadline expired on 12 October 2009.


Solution. We will use the well-known fact that \nu(n) = \sum_{k=1}^\infty \bigg[\frac{n}{2^k}\bigg].

If the base-2 form of the number n is n=\overline{d_\ell
  d_{\ell-1}\ldots d_1d_0} = \sum\limits_{i=0}^\ell d_i\cdot2^i, where the digits d_0,\ldots,d_\ell are all 0 or 1, then


  \nu(n) =
  \sum_{k=1}^\infty \bigg[\frac{n}{2^k}\bigg] =
  \sum_{k=1}^\ell \Bigg[\frac{\overline{d_\ell d_{\ell-1}\ldots d_1d_0}}{2^k}\Bigg] =
  \sum_{k=1}^\ell \overline{d_\ell d_{\ell-1}\ldots d_k} =


  = \sum_{k=1}^\ell \left(\sum_{i=k}^\ell d_i \cdot 2^{i-k}\right) =
  \sum_{i=1}^\ell d_i \left(\sum_{k=1}^i 2^{i-k}\right) =
  \sum_{i=1}^\ell d_i \cdot (2^i-1).

Now we will show that there exists an integer r which is relatively prime to m, and an infinite sequence i_1<i_2<i_3<\ldots of positive integers such that


2^{i_1}-1 \equiv 2^{i_2}-1 \equiv 2^{i_3}-1 \equiv 
\ldots \equiv r \pmod{m}.

Let m=2tu where u is odd, and consider an arbitrary positive integer i\get for which \varphi(u) divides i-1. By the Euler-Fermat theorem,


u \,\Big|\, 2^{\varphi (u)}-1 \,\Big|\, 2^{i-1}-1,


2^i-1 = 2(2^{i-1}-1) + 1 \equiv 1 \pmod{u},
(1)

and, due to i\get,


2^i-1 \equiv -1 \pmod{2^t}.
(2)

The relations (1) and (2) determine the residue class of 2i-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 a\equiv ur\pmod{m}. For n=2^{i_1}+\ldots+2^{i_u} we achieve


\nu(n) =
\nu\big( 2^{i_1}+\ldots+2^{i_u} \big) =
(2^{i_1}-1)+\ldots+(2^{i_u}-1) \equiv
u\cdot r \equiv a \pmod{m}.

Based on the solution of András Éles


Statistics on problem A. 486.
13 students sent a solution.
5 points:Ágoston Tamás, Backhausz Tibor, Bodor Bertalan, Éles András, Frankl Nóra, Nagy 235 János, Nagy 648 Donát, Somogyi Ákos, Strenner Péter, Szabó 928 Attila.
4 points:Kalina Kende.
3 points:1 student.
0 point:1 student.


  • Problems in Mathematics of KöMaL, September 2009

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