Olimpiadi della Matematica

Solution:

The answer is $(\mathbf{D})$. Let $p(x)$ be the polynomial; $p(1)$ is the sum of all the coefficients and $p(-1)$ is the alternating sum of the coefficients, those of even index with a positive sign and those of odd index with a negative sign. Consequently, $p(1)+p(-1)$ is twice the sum of the coefficients of even index, while $p(1)-p(-1)$ is twice the sum of the coefficients of odd index.

However, we observe that in $M$ there appear, as coefficients of the $a_{k}$ of even index, the powers $2^{2 k}$, while as coefficient of $a_{2 k+1}$ there appears $3^{2 k}$.

Let us then consider
$$\begin{aligned} & p(2)+p(-2)=2\left(a_{0}+2^{2} a_{2}+\ldots+2^{98} a_{98}+2^{100} a_{100}\right) \\ & p(3)-p(-3)=2\left(3 a_{1}+3^{3} a_{3}+\ldots+3^{97} a_{97}+3^{99} a_{99}\right) . \end{aligned}$$
Hence
$$\begin{aligned} M & =\frac{1}{2}(p(2)+p(-2))-\frac{1}{6}(p(3)-p(-3)) \\ & =\frac{1}{2}\left((-5)^{50}+(-25)^{50}\right)-\frac{1}{6}\left(5^{50}-(-25)^{50}\right) \\ & =5^{50}\left(\frac{5^{50}+1}{2}+\frac{5^{50}-1}{6}\right) . \end{aligned}$$
At this point we have factored out $5^{50}$, and to conclude it suffices to observe that the quantity in parentheses is an integer ($5^{50}+1$ 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).