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2010 AMC 10B Problems/Problem 25

Revision as of 16:17, 14 October 2016 by Pega969 (talk | contribs) (→‎Critique (cont.))

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.

Critique of Critique

First of all, the solution shows that $a$ is a multiple of $315$, not a factor of $315$. Many people confuse the usage of the words 'factor' and 'multiple'. Secondly, even if $a$ is a multiple of $315$, it does not mean that you can instantly get that the answer is $315$ because we need to know that $a=315$ is possible. After all, $a$ is also a multiple of $1$, but $1$ is definitely not the smallest possible number.

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

\[P(x) = (x-1)(x-3)(x-5)(x-7)((-8x + 60)(x-2)(x-6)+42) + 315\] \[= -8 x^7+252 x^6-3248 x^5+22050 x^4-84392 x^3+179928 x^2-194592 x+80325\]

Critique of Critique

The problem states the "least value" of $a$, so it is not needed to add the extra steps.

Critique of Critique of Critique

It is still necessary to show that the minimum is achievable. For example $x^2+1>0$, but $0$ is not the least value of $x^2+1$

See also

2010 AMC 10B (Problems • Answer Key • Resources)
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Problem 24 		Followed by
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All AMC 10 Problems and Solutions

Seealso The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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