11th APMO 1999 Problem 1 Find the smallest positive integer n such that no arithmetic progression of 1999 reals contains just n integers. Solution
: 70. Using a difference of 1/n , where n does not divide 1999, we can get a progression of 1999 terms with m = [1998/n] or m = [1998/n] - 1 integers. Thus {0, 1/n, 2/n, ... , 1998/n} has m+1 integers, and {1/n, 2/n, ... , 1999/n} has m integers. So we are ok until n gets so large that the gap between [1998/n] and [1998/(n+1)] is 3 or more. This could happen for 1998/n - 1998/(n+1) just over 2 or n > 31. So checking, we find [1998/31] = 64, [1998/30] = 66, [1998/29] = 68, [1998/28] = 71. We can get 68 integers with {1/29, 2/29, ... , 1999/29} and 69 with {0, 1/29, 2/29, ... , 1998/29}. We can get 71 with {1/28, 2/28, ... , 1999/28}, but we cannot get 70. Note that a progression with irrational difference gives at most 1 integer. A progression with difference a/b, where a and b are coprime integers, gives the same number of integers as a progression with difference 1/b. So it does not help to widen the class of progressions we are looking at. 11th APMO 1999 © John Scholes jscholes@kalva.demon.co.uk 11 Apr 2002