HMMT 十一月 2009 · 冲刺赛 · 第 29 题
HMMT November 2009 — Guts Round — Problem 29
题目详情
- [ 17 ] For how many integer values of b does there exist a polynomial function with integer coefficients such that f (2) = 2010 and f ( b ) = 8?
解析
- [ 17 ] For how many integer values of b does there exist a polynomial function with integer coefficients such that f (2) = 2010 and f ( b ) = 8? 2002 Answer: 32 We can take f ( x ) = − ( x − b ) + 2010 for all divisors d of − 2002. To see that we d can’t get any others, note that b − 2 must divide f ( b ) − f (2), so b − 2 divides − 2002 (this is because n n n n b − 2 divides b − 2 and hence any sum of numbers of the form b − 2 ).