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

Revision as of 23:17, 21 April 2010

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

$P(2)-a=(2-1)(2-3)(2-5)(2-7)Q(2) = -15Q(2) = -2a$ $P(4)-a=(4-1)(4-3)(4-5)(4-7)Q(4) = -15Q(4) = -2a$ $P(6)-a=(6-1)(6-3)(6-5)(6-7)Q(6) = -15Q(6) = -2a$ $P(8)-a=(8-1)(8-3)(8-5)(8-7)Q(8) = -15Q(8) = -2a$

$-2a=-15Q(2)=9Q(4)=-15Q(6)=105Q(8).$ Thus, the least value of $a$ must be the $lcm(15,9,15,105)$ . Solving, we receive $315$ , so our answer is $\boxed{\textbf{(B)}\ 315}$ .

@@ Line 1: / Line 1: @@
-P(x)-a=(x-1)(x-3)(x-5)(x-7)Q(x).
+There must be some polynomial <math>Q(x)</math> such that <math>P(x)-a=(x-1)(x-3)(x-5)(x-7)Q(x)</math>
-Then, plugging in values of 2,4,6,8, we get
--2a=-15Q(2)=9Q(4)=-15Q(6)=105Q(8).
+Then, plugging in values of <math>2,4,6,8,</math> we get
-Thus, the least value of a must be the lcm(15,9,15,105).
-Solving, we receive 315, so a=15.
+<math>P(2)-a=(2-1)(2-3)(2-5)(2-7)Q(2) = -15Q(2) = -2a</math>
+<math>P(4)-a=(4-1)(4-3)(4-5)(4-7)Q(4) = -15Q(4) = -2a</math>
+<math>P(6)-a=(6-1)(6-3)(6-5)(6-7)Q(6) = -15Q(6) = -2a</math>
+<math>P(8)-a=(8-1)(8-3)(8-5)(8-7)Q(8) = -15Q(8) = -2a</math>
+<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>.
+Solving, we receive <math>315</math>, so our answer is <math> \boxed{\textbf{(B)}\ 315} </math>.

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

Revision as of 23:17, 21 April 2010