# Difference between revisions of "2010 AMC 10B Problems/Problem 25"

## Problem

Let $a > 0$, and let $P(x)$ be a polynomial with integer coefficients such that

$P(1) = P(3) = P(5) = P(7) = a$, and
$P(2) = P(4) = P(6) = P(8) = -a$.

What is the smallest possible value of $a$?

$\textbf{(A)}\ 105 \qquad \textbf{(B)}\ 315 \qquad \textbf{(C)}\ 945 \qquad \textbf{(D)}\ 7! \qquad \textbf{(E)}\ 8!$

## Solution

We observe that because $P(1) = P(3) = P(5) = P(7) = a$, if we define a new polynomial $R(x)$ such that $R(x) = P(x) - a$, $R(x)$ has roots when $P(x) = a$; namely, when $x=1,3,5,7$.

Thus since $R(x)$ has roots when $x=1,3,5,7$, we can factor the product $(x-1)(x-3)(x-5)(x-7)$ out of $R(x)$ to obtain a new polynomial $Q(x)$ such that $(x-1)(x-3)(x-5)(x-7)(Q(x)) = R(x) = P(x) - a$.

Then, plugging in values of $2,4,6,8,$ we get

$$P(2)-a=(2-1)(2-3)(2-5)(2-7)Q(2) = -15Q(2) = -2a$$ $$P(4)-a=(4-1)(4-3)(4-5)(4-7)Q(4) = 9Q(4) = -2a$$ $$P(6)-a=(6-1)(6-3)(6-5)(6-7)Q(6) = -15Q(6) = -2a$$ $$P(8)-a=(8-1)(8-3)(8-5)(8-7)Q(8) = 105Q(8) = -2a$$

$-2a=-15Q(2)=9Q(4)=-15Q(6)=105Q(8).$ Thus, the least value of $a$ must be the $lcm(15,9,15,105)$. Solving, we receive $315$, so our answer is $\boxed{\textbf{(B)}\ 315}$.

## Critique

The above solution is incomplete. What is really proven is that 315 is a factor of $a$, if such an $a$ exists. That only rules out answer A.

To prove that the answer is correct, one could exhibit a polynomium that satisfies the requirements with $a=315$. Here's one: $P(x) = -8x^7 + 252 x^6 -3248x^5 + 22050x^4 -84392x^3 + 179928x^2-194592x+80325$.

You get that from the $8\times8$ matrix $M_{i,j} = i^{j-1}$ and $y^T=(315,-315,\ldots,315,-315)$ and computing $c=M^{-1}y$ which comes out as the all-integer coefficients above.