2010 AMC 12B Problems/Problem 21

Problem 21

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 $\text{lcm}(15,9,15,105)$ . Solving, we receive $315$ , so our answer is $\boxed{\textbf{(B)}\ 315}$ .

To complete the solution, we can let $a = 315$ , and then try to find $Q(x)$ . We know from the above calculation that $Q(2)=42, Q(4)=-70, Q(6)=42$ , and $Q(8)=-6$ . Then we can let $Q(x) = T(x)(x-2)(x-6)+42$ , getting $T(4)=28, T(8)=-4$ . Let $T(x)=L(x)(x-8)-4$ , then $L(4)=-8$ . Therefore, it is possible to choose $T(x) = -8(x-8)-4 = -8x + 60$ , so the goal is accomplished. As a reference, the polynomial we get is

$P(x) = (x-1)(x-3)(x-5)(x-7)((-8x + 60)(x-2)(x-6)+42) + 315$ $= -8 x^7+252 x^6-3248 x^5+22050 x^4-84392 x^3+179928 x^2-194592 x+80325$ The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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2010 AMC 12B Problems/Problem 21

Problem 21

Solution