2010 AMC 12B Problem 21

Below is the professionally curated solution for Problem 21 of the 2010 AMC 12B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2010 AMC 12B solutions, or check the answer key.

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Concepts:polynomial divisibility least common multiple

Difficulty rating: 2300

21.

Let $a\gt0,$ 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 $a?$

$105$

$315$

$945$

$7!$

$8!$

Solution:

Since $1, 3, 5, 7$ are roots of $P(x)-a,$ write $P(x)-a=(x-1)(x-3)(x-5)(x-7)Q(x)$ with $Q$ having integer coefficients.

Evaluating at $x=2, 4, 6, 8$ (where $P=-a$ ) gives $-2a=-15\,Q(2)=9\,Q(4)=-15\,Q(6)=105\,Q(8).$

So $15, 9,$ and $105$ all divide $2a,$ hence $\operatorname{lcm}(15,9,105)=315$ divides $2a.$ Since $315$ is odd, $315\mid a,$ so $a\ge315.$

The value $a=315$ is attainable, so the smallest is $315.$

Thus, the correct answer is B.

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