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.