32nd IMO 1991 shortlist Problem 18 What is the largest power of 1991 which divides 1990 m + 1992 n , where m = 1991 1992 and n = 1991 1990 ?
1991 1991 Solution We show first that for odd h > 1 and k > 0, we have (1 + h) H = 1 + s k h k+1 , where H = h k and h does not divide s k (in other words h k+1 is the highest power of h dividing (1 + h) H - 1). We use induction on k. For k = 1, we have (by the binomial theorem): (1 + h) h = 1 + h·h + ½(h-1)h 3 + terms in h 3 and higher = 1 + h 2 (1 + bh). So the result is true for k = 1. Suppose it is true for k. Then taking H' = h k+1 , we have (1 + h) H' = (1 + s k h k+1 ) h = 1 + h s k h k+1 + ½(h-1) s k 2 h 2k+3 + terms higher than h k+2 = 1 + h k+2 (s k + hb'), so it is true for k+1, and hence for all k. An exactly similar argument shows that h k+1 is the highest power of h dividing (h - 1) H + 1. Thus 1991 1991 is the highest power of 1991 dividing 1992 n - 1 and 1991 1993 is the highest power of 1991 dividing 1990 m + 1. Hence 1991 1991 is the highest power of 1991 dividing their sum. 32nd IMO shortlist 1991 © John Scholes jscholes@kalva.demon.co.uk 1 Jan 2003 Last corrected/updated 1 Jan 03