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The Problem Solving Competition–November 2012 Problem A: Polynomial of which degree? Determine all positive integers n for which there is a polynomial P (x) of degree n, satisfying all of the following conditions: (1) P (k) = k for k = 1, 2, · · · , n; (2) P (0) is an integer; (3) P (−1) = 2012. Solution: Let Q(x) = P (x) − x. By (1), Q(x) = C(x − 1)(x − 2) · · · (x − n) for some n (−1) . By (2), Q(0) = constant C. By (3), C = 2013 (n+1)! 2013 n+1 must be an integer. Since 2013 = 3 · 11 · 61, the possible divisors of 2013 are 1, 3, 11, 33, 61, 183, 671 and 2013. Hence n = 2, 10, 32, 60, 182, 670, 2012. Problem B: Sequence Let x0 = 0, x1 = 1, and for n ≥ 1 xn+1 = 1 1 xn + (1 − )xn−1 n+1 n+1 Determine lim xn . n→∞ Solution: We know xn+1 (n + 1) = xn + nxn−1 , and then (n + 1)(xn+1 − xn ) = −n(xn − −n xn−1 ). Let yn = xn+1 − xn , then (n + 1)y − n = −nyn−1 . So yn = yn−1 . Then n+1 n−1 X −n −(n − 1) −1 (−1)n yn = · ··· y0 = since y0 = x1 − x0 = 1. Hence xn = x0 + yi = n+1 n 2 n+1 i=1 n−1 ∞ X X (−1)i (−1)i . It follows that lim xn = = ln 2 by the Taylor series for ln(1 + x) n→∞ i + 1 i + 1 i=0 i=0 evaluated at 1.