2010 AMC 10B Problem 25

Because $1,$ $3,$ $5,$ and $7$ are roots of $P(x)-a$, write $P(x)-a=(x-1)(x-3)$ $\cdot(x-5)(x-7)Q(x)$, where $Q(x)$ has integer coefficients. Substituting $2,$ $4,$ $6,$ and $8$ for $x$ gives $-2a=-15Q(2)$ $=9Q(4)$ $=-15Q(6)$ $=105Q(8)$. Hence $a$ must be a multiple of $\operatorname{lcm}(15,9,105)=315$. This lower bound is attainable: take $Q(x)=42$ $+(x-2)(x-6)(60-8x)$ and define $P(x)=315$ $+(x-1)(x-3)$ $\cdot(x-5)(x-7)Q(x)$. This polynomial has integer coefficients and satisfies the required values. Thus, B is the correct answer.