Problem:
For certain pairs (m,n) of positive integers with m≥n there are exactly 50 distinct positive integers k such that ∣logm−logk∣<logn. Find the sum of all possible values of the product mn.
Solution:
The inequality ∣logm−logk∣<logn is equivalent to −logn<logm− logk<logn, which is equivalent to lognm<logk<logmn. Write m=nq+r, where q is a positive integer and r is an integer with 0≤r<n. The inequality then becomes
log(q+nr)<logk<log(n(nq+r))
There are n(nq+r)−q−1 possible values of k, namely, q+1,q+2,…, n(nq+r)−1. By the given condition, n(nq+r)−q−1=50 or (n2−1)q+nr=51. The potential values of n are 2,3,…,7. The only solutions are (n,q,r)=(2,17,0) and (3,6,1). Hence (m,n)=(34,2) or (19,3), and mn=68 or 57. Thus the sum is 125.
The problems on this page are the property of the MAA's American Mathematics Competitions