12. Given the polynomial

$$
\left(x^{2}+5 x-19\right)^{50}=a_{0}+a_{1} x+a_{2} x^{2}+\cdots+a_{100} x^{100},
$$

consider the quantity

$$
M=a_{0}-a_{1}+4 a_{2}-9 a_{3}+16 a_{4}-81 a_{5}+64 a_{6}-729 a_{7}+\cdots+2^{98} a_{98}-3^{98} a_{99}+2^{100} a_{100}
$$

where the coefficient of $a_{2 k}$ is $2^{2 k}$ and the coefficient of $a_{2 k+1}$ is $-3^{2 k}$. What is the highest power of 5 that divides $M$?
(A) $5^{0}$
(B) $5^{1}$
(C) $5^{5}$
(D) $5^{50}$
(E) $5^{100}$

## Numerical Answer Problems -5 points
12. The answer is $(\mathbf{D})$. Let $p(x)$ be the polynomial. $p(1)$ is the sum of all coefficients, and $p(-1)$ is the sum of the coefficients with alternating signs, where the coefficients of even indices are positive and those of odd indices are negative. Consequently, $p(1)+p(-1)$ is twice the sum of the coefficients of even indices, while $p(1)-p(-1)$ is twice the sum of the coefficients of odd indices.

However, in $M$, the coefficients of $a_{k}$ of even indices are the powers $2^{2 k}$, while the coefficient of $a_{2 k+1}$ is $3^{2 k}$.

Consider then

$$
\begin{aligned}
& p(2)+p(-2)=2\left(a_{0}+2^{2} a_{2}+\ldots+2^{98} a_{98}+2^{100} a_{100}\right) \\
& p(3)-p(-3)=2\left(3 a_{1}+3^{3} a_{3}+\ldots+3^{97} a_{97}+3^{99} a_{99}\right) .
\end{aligned}
$$

Thus,

$$
\begin{aligned}
M & =\frac{1}{2}(p(2)+p(-2))-\frac{1}{6}(p(3)-p(-3)) \\
& =\frac{1}{2}\left((-5)^{50}+(-25)^{50}\right)-\frac{1}{6}\left(5^{50}-(-25)^{50}\right) \\
& =5^{50}\left(\frac{5^{50}+1}{2}+\frac{5^{50}-1}{6}\right) .
\end{aligned}
$$

At this point, we have factored out a $5^{50}$, and to conclude, we just need to observe that the quantity in parentheses is an integer (since $5^{50}+1$ is even, as is $5^{50}-1$, which is also a multiple of 3) and is not a multiple of 5 (because $4 \cdot 5^{50}+2$ is not a multiple of 5).