Putnam 1993 Problem B1 What is the smallest integer n > 0 such that for any integer m in the range 1, 2, 3, ... , 1992 we can always find an integral multiple of 1/n in the open interval (m/1993, (m + 1)/1994)? Solution
: 1993 + 1994 = 3987. Easy. For positive a, b, a', b', with a/b  k'/n > 1/1994, where k' = n - k, or 1993k' < n < 1994k'. That is clearly not possible for k' = 1, so the smallest possible n corresponds to k' = 2 and is 3987. Putnam 1993 © John Scholes jscholes@kalva.demon.co.uk 12 Dec 1998