11th Swedish 1971 Problem 4 Find (65533 3 + 65534 3 + 65535 3 + 65536 3 + 65537 3 + 65538 3 + 65539 3 )/(32765·32766 + 32767·32768 + 32768·32769 + 32770·32771).
7·2 16 Solution Put N = 65536 = 2 16 . Then numerator = (N-3) 3 + (N-2) 3 + (N-1) 3 + N 3 + (N+1) 3 + (N+2) 3 + (N+3) 3 = ((N-3) 3 + (N+3) 3 ) + ((N-2) 3 + (N+2) 3 ) + ((N-1) 3 + (N+1) 3 ) + N 3 = 2(N 3 + 27N) + 2(N 3 + 12N) + 2(N 3 + 3N) + N 3 = 7(N 3 + 12N). Put M = N/2 = 32768. Denominator = (M-3)(M-2) + (M-1)M + M(M+1) + (M+2)(M+3) = 4M 2 + 12 = N 2 + 12. Hence expr = 7N. 11th Swedish 1971 © John Scholes jscholes@kalva.demon.co.uk 30 December 2003 Last corrected/updated 30 Dec 03