46th Putnam 1985 Problem B2 The polynomials p n (x) are defined as follows: p 0 (x) = 1; p n+1 '(x) = (n + 1) p n (x + 1), p n+1 (0) = 0 for n ≥ 0. Factorize p 100 (1) into distinct primes. Solution
: (101) 99 . We show that p n (x) = x (x + n) n-1 for n > 0. In fact, this is an easy induction. We have p n+1 '(x) = (n + 1) (x + n + 1) n - n(n + 1)(x + n + 1) n-1 . Integrating: p n+1 (x) = (x + n + 1) n+1 - (n + 1)(x + n + 1) n = x(x + n + 1) n . 46th Putnam 1985 © John Scholes jscholes@kalva.demon.co.uk 7 Jan 2001