2010 AMC 12B Problem 21

Question

Let

a\gt0,

and let

P(x)

be a polynomial with integer coefficients such that

\begin{aligned} &P(1)=P(3)=P(5) \\ &\quad {}=P(7)=a, \end{aligned}

and

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

What is the smallest possible value of

a?

105

315

945

7!

8!

Answer: B

Concepts:polynomial divisibility least common multiple

Difficulty rating: 2300

Small hint:

P(x)-a=(x-1)(x-3)

\cdot(x-5)(x-7)Q(x)

for some integer polynomial

Q

Big hint:

Evaluate at

x=2,4,6,8

to force

2a

to be divisible by

\operatorname{lcm}(15,9,105)

Solution:

Since

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)

with

Q

having integer coefficients.

Evaluating at

x=2, 4, 6, 8

(where

P=-a

) gives

\begin{aligned} -2a &=-15\,Q(2) \\ &=9\,Q(4) \\ &=-15\,Q(6) \\ &=105\,Q(8). \end{aligned}

So

15, 9,

and

105

all divide

2a,

hence

\operatorname{lcm}(15,9,105)=315

divides

2a.

Since

315

is odd,

a

is divisible by

315,

so

a\ge315.

To attain the bound, let

\begin{aligned} F(x)&=(x-1)(x-3) \\ &\quad\cdot(x-5)(x-7), \\ Q(x)&=42+(x-2)(x-6) \\ &\quad\cdot(60-8x), \\ P(x)&=315+F(x)Q(x). \end{aligned}

This integer polynomial equals

315

at

1,3,5,7.

At

2,4,

the pairs

(F(x),Q(x))

are

(-15,42),(9,-70),

and at

6,8

they are

(-15,42),(105,-6).

Thus

F(x)Q(x)=-630

each time and

P(x)=-315.

Hence the bound is attainable.

Thus, the correct answer is B.