2010 AMC 10B Problem 25

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

\begin{aligned} P(2)&=P(4)=P(6)\\ &=P(8)=-a. \end{aligned}

What is the smallest possible value of

a?

105

315

945

7!

8!

Use the roots of

P(x)-a

Plug

2,

4,

6,

and

8

into the factored form to force divisibility of

a

Because

1,

3,

5,

and

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

2,

4,

6,

and

8

for

x

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.

Problem 25 in other years