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

Revision as of 20:07, 24 January 2016

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 the (Critique of the 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

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

@@ Line 34: / Line 34: @@
 You get that from the <math>8\times8</math> matrix <math>M_{i,j} = i^{j-1}</math> and <math>y^T=(315,-315,\ldots,315,-315)</math> and computing <math>c=M^{-1}y</math> which comes out as the all-integer coefficients above.
-== Critique of the Critique ==
-Once you find that <math>a</math> is a factor of 315, you can instantly find that the answer is 315 because the problem asks for the smallest value of <math>a</math>.
 == Critique of the (Critique of the Critique)==

2010 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 24	Followed by Last question
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

Page

Toolbox

Search