2010 AMC 10B Problems/Problem 25

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

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2010 AMC 10B Problems/Problem 25