POLYA013:  Length of the nonperiodic part of the decimal expansion of 1/N (proposed by Luis Lopes,  11/19/01).

Let N be a whole number. The decimal expansion of  1/N is periodic after a certain number of digits beyond the decimal point. We call this the length of  N, or more precisely, the length of the nonperiodic part of the decimal expansion of 1/N.
For example,  11, 25, 55 have lengths 0, 2, 1 respectively:

1/11 = 0.09090909...
1/36 = 0.027777...
1/55 = 0.0181818...

Show that the length of N is the larger of the exponents of 2 and 5 in the prime decomposition of N, i.e., if
N = 2^(a_2) 3^(a_3) 5^(a^5) ..., then the length of N is max (a_2, a_5).

Discussion. [FvL, 11/21/01]: Let L'=max(a_2,a_5). Since 1/N has a periodic part of its decimal expansion, there is a smallest number M such that (10^M - 1)/N has terminating decimal expansion. The number of decimals of (10^M - 1)/N is equal to the length L of N. Hence, for an integer P, we have (10^(M+L) - 10^L)/N = P where P is not a positive multiple of 10, or 10^(M+L) - NP = 10^L. So 10^L'=10^L.

Bibliography.