11. Let $p(x)$ be a polynomial with integer coefficients such that $p(0)=6$. It is known that among the integers $m$ between 1 and 60, exactly 40 are such that $p(m)$ is a multiple of 3; furthermore, it is known that among the integers $m$ between 1 and 60, exactly 30 are such that $p(m)$ is a multiple of 4. How many integers $m$ between 1 and 60 are such that $p(m)$ is a multiple of 6? Note: all intervals that appear in this problem are to be considered inclusive.
(A) 10
(B) 20
(C) 25
(D) 30
(E) 40
11. The answer is $\mathbf{( E )}$. To prove this, we want to show that $p(m)$ is even for every integer $m$, and that consequently $p(m)$ is a multiple of 6 if and only if it is a multiple of 3. At this point, we can use the hypothesis that the integers $m$ for which $p(m)$ is a multiple of 3 are exactly 40 to obtain the answer.

First, observe that the parity of $p(m)$ depends only on the parity of $m$: indeed, the parity of a monomial $a \cdot m^{k}$ depends only on the parities of $m$ and $a$, and the parity of the value of $p(m)$, which is a sum of monomials in $m$, depends only on the parity of each term. It is therefore sufficient to show that $p(m)$ is even for at least one even value and for at least one odd value of $m$. By hypothesis, $p(0)=6$, so we need to show the existence of an odd value of $m$ for which $p(m)$ is even. We can write $p(x)=x \cdot q(x)+6$, where $q(x)$ is a polynomial with integer coefficients. If $m$ is a multiple of 4, say $m=4 k$, then $p(m)=4 k \cdot q(4 k)+6$ is not a multiple of 4 (it gives a remainder of 2 when divided by 4). By hypothesis, there are 30 integers between 1 and 60 such that $p(m)$ is a multiple of 4; since the $m$ that are multiples of 4 do not have this property, and the $m$ that are even but not multiples of 4 are only 15, there must exist odd integers such that $p(m)$ is even.