Question

Let $a \gt 0,$ and let $P(x)$ be a polynomial with integer coefficients such that $\begin{aligned} P(1) &= P(3) \\ &= P(5) = P(7) = a, \end{aligned}$ and $P(2) = P(4) = P(6) = P(8)$ $= -a.$ What is the smallest possible value of $a?$

$105$

$315$

$945$

$7!$

$8!$

Answer: B

Concepts:polynomial divisibility least common multiple

Difficulty rating: 2350

Solution:

Because $1,3,5,7$ are roots of $P(x)-a$ , write $P(x)-a=(x-1)(x-3)$ $\cdot(x-5)(x-7)Q(x)$ , where $Q(x)$ has integer coefficients.

Substituting $x=2,4,6,8$ gives $-2a=-15Q(2)$ $=9Q(4)$ $=-15Q(6)$ $=105Q(8)$ . Hence $a$ must be a multiple of $\operatorname{lcm}(15,9,105)=315$ .

This lower bound is attainable: take $Q(x)=42$ $+(x-2)(x-6)(60-8x)$ and define $P(x)=315$ $+(x-1)(x-3)$ $\cdot(x-5)(x-7)Q(x)$ . This polynomial has integer coefficients and satisfies the required values.

Thus, B is the correct answer.

2010 AMC 10B Problem 25

Answer: B

Difficulty rating: 2350

Solution:

Problem 25 in Other Years