10. Let $N$ be the maximum number obtainable as the product of distinct positive integers whose sum is 2024. How many "0" digits does the decimal representation of $N$ end with?
(A) 9
(B) 10
(C) 12
(D) 13
(E) 14
10. The answer is (D). Let $A$ be the set of numbers that sum to 2024 and whose product is the maximum possible, i.e., $N$. I call "hole" a number $n$ that is not in $A$, but such that $A$ contains both a number greater than $n$ and a number less than $n$.

I prove that $A$ does not contain 1: let $M$ be the maximum integer in $A$; if $A$ contained 1, then the set $B$ obtained from $A$ by removing 1 and adding $M+1$ would still satisfy all the conditions imposed by the problem, but the product of the numbers in it would be $N \cdot (M+1) > N$, contradicting the fact that $N$ was the maximum possible.

I prove by contradiction that $A$ has at most one hole: suppose that $a < b$ are two holes in $A$; since $a < b$, the set $C$ obtained from $A$ by removing $a, b$ and adding $a+1$ and $b-1$ still satisfies all the conditions of the problem. Moreover, the product of the numbers in $C$ would be given by $N \frac{(a+1)(b-1)}{a b} > N$, contradicting the maximality of $N$.

Furthermore, $A$ cannot not contain 2 and at the same time have a hole $n$ such that $n+2 \in A$. Indeed, if this were the case, the set $D$, obtained from $A$ by adding 2 and $n$ and removing $n+2$, would still satisfy the conditions of the problem. Once again, the product of the numbers in $D$ is given by $N \frac{2 n}{n+2} > N$, implying that $N$ was not the maximum possible.

The only way to write 2024 as the sum of distinct numbers while respecting the previous conditions is to use the numbers from 2 to 64, skipping 55. It follows that $N = \frac{64!}{55}$, which ends with 13 zeros. To count the number of trailing zeros, it is sufficient to count the number of factors of 2 and factors of 5 that appear in the factorization of $N$, and then take the minimum of these two quantities. It is easy to verify that there are fewer factors of 5 than factors of 2; finally, to calculate the number of factors of 5, one can observe that these are the same as in the product $5 \cdot 10 \cdot 15 \cdots 50 \cdot 60$: this product has $1+1+1+1+2+1+1+1+1+2+1=13$ factors of 5 (indeed, each multiple of 5 between 5 and 60 contains one factor of 5, except for 25 and 50 which contain 2).