An integrality proof by Legendre's formula

1. The number of trailing zeros is $\min(v_2,v_5)$ of $2024!$, and $v_5\le v_2$ because $\lfloor 2024/5^k\rfloor\le\lfloor 2024/2^k\rfloor$ for every $k$. So
$$v_5(2024!)=\left\lfloor\frac{2024}5\right\rfloor+\left\lfloor\frac{2024}{25}\right\rfloor+\left\lfloor\frac{2024}{125}\right\rfloor+\left\lfloor\frac{2024}{625}\right\rfloor=404+80+16+3=503.$$

2. Floor inequality. Write $x=a+\alpha$, $y=b+\beta$ with $a,b\in\mathbb{Z}$, $\alpha,\beta\in[0,1)$. The inequality becomes
$$\lfloor2\alpha\rfloor+\lfloor2\beta\rfloor\ge\lfloor\alpha+\beta\rfloor.$$
If $\alpha+\beta<1$ the right side is $0$. If $\alpha+\beta\ge1$ then one of $\alpha,\beta$ is at least $\frac12$, so the left side is at least $1$, and the right side is exactly $1$ (since $\alpha+\beta<2$).

Integrality. For each prime $p$, by Legendre's formula,
$$v_p(\text{expression})=\sum_{k\ge1}\left(\left\lfloor\frac{2m}{p^k}\right\rfloor+\left\lfloor\frac{2n}{p^k}\right\rfloor-\left\lfloor\frac m{p^k}\right\rfloor-\left\lfloor\frac n{p^k}\right\rfloor-\left\lfloor\frac{m+n}{p^k}\right\rfloor\right)\ge0$$
by the inequality with $x=\frac m{p^k}$, $y=\frac n{p^k}$. All the valuations are non-negative, so the number is an integer.

Check. $(m,n)=(1,1)$: $\frac{2\cdot2}{1\cdot1\cdot2}=2$; $(2,3)$: $\frac{24\cdot720}{2\cdot6\cdot120}=12$ ✓.

Remark. These are the "super Catalan numbers". The case $m=n$ gives $\binom{2n}n$.

Lesson. To prove that a quotient of factorials is an integer, compare prime by prime using Legendre's formula, and reduce to an elementary inequality for floor functions.