Difference between revisions of "2010 AMC 12B Problems/Problem 21"

Revision as of 14:11, 7 February 2012

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

There must be some polynomial $Q(x)$ such that $P(x)-a=(x-1)(x-3)(x-5)(x-7)Q(x)$

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 $\boxed{\textbf{(B)}\ 315}$ .

2010 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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 12 Problems and Solutions

@@ Line 25: / Line 25: @@
 <math>-2a=-15Q(2)=9Q(4)=-15Q(6)=105Q(8).</math>
-Thus, the least value of <math>a</math> must be the <math>lcm(15,9,15,105)</math>.
+Thus, the least value of <math>a</math> must be the <math>\text{lcm}(15,9,15,105)</math>.
 Solving, we receive <math>315</math>, so our answer is <math> \boxed{\textbf{(B)}\ 315} </math>.
 == See also ==
 {{AMC12 box|year=2010|num-b=20|num-a=22|ab=B}}

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Difference between revisions of "2010 AMC 12B Problems/Problem 21"

Revision as of 14:11, 7 February 2012

Problem 21

Solution

See also