AMC10 2010 B

Answer (B): Because 1, 3, 5, and 7 are roots of the polynomial $P(x)-a$, it follows that $P(x)-a=(x-1)(x-3)(x-5)(x-7)Q(x),$ where $Q(x)$ is a polynomial with integer coefficients. The previous identity must hold for $x=2,4,6,$ and $8,$ thus $-2a=-15Q(2)=9Q(4)=-15Q(6)=105Q(8).$ Therefore $315=\operatorname{lcm}(15,9,105)$ divides $a$, that is $a$ is an integer multiple of 315. Let $a=315A$. Because $Q(2)=Q(6)=42A$, it follows that $Q(x)-42A=(x-2)(x-6)R(x)$ where $R(x)$ is a polynomial with integer coefficients. Because $Q(4)=-70A$ and $Q(8)=-6A$ it follows that $-112A=-4R(4)$ and $-48A=12R(8)$, that is $R(4)=28A$ and $R(8)=-4A$. Thus $R(x)=28A+(x-4)(-6A+(x-8)T(x))$ where $T(x)$ is a polynomial with integer coefficients. Moreover, for any polynomial $T(x)$ and any integer $A$, the polynomial $P(x)$ constructed this way satisfies the required conditions. The required minimum is obtained when $A=1$ and so $a=315$.