6th Chinese 1991 Problem B2 Let f(n) be the smallest of 1 + [n/1], 4 + [n/4], 9 + [n/9], 16 + [n/16], 25 + [n/25], ... . For which n is f(n) = 1991? Solution
: n = 990208, 990209, 990210, ... , 991231. f(n) = 1991 iff (1) 1991k 2 - k 4 ≤ n for all k, and (2) 1992k 2 - k 4 > n for some k. (1) is equivalent to (1991/2) 2 - (k 2 - 1991/2) 2 = 990208. (2) is equivalent to 996 2 - (k 2 - 996) 2 > n for some k. The smallest value of (k 2 - 996) is 28, so the smallest value of 996 2 - (k 2 - 996) 2 is 996 2 - 28 2 = 991232. Hence (2) is equivalent to n < 991232. 6th Chinese 1991 © John Scholes jscholes@kalva.demon.co.uk 1 May 2002