32nd IMO shortlist 1991/18 solution

32nd IMO 1991 shortlist

Problem 18

What is the largest power of 1991 which divides 1990^m + 1992ⁿ, where m = 1991¹⁹⁹² and n = 1991¹⁹⁹⁰?

Answer

1991¹⁹⁹¹

Solution

We show first that for odd h > 1 and k > 0, we have (1 + h)^H = 1 + s_kh^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)h³ + terms in h³ and higher = 1 + h²(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_kh^k+1)^h = 1 + h s_kh^k+1 + ½(h-1) s_k²h^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 1991¹⁹⁹¹ is the highest power of 1991 dividing 1992ⁿ - 1 and 1991¹⁹⁹³ is the highest power of 1991 dividing 1990^m + 1. Hence 1991¹⁹⁹¹ is the highest power of 1991 dividing their sum.

32nd IMO shortlist 1991