[Number Theory]
Problem: Given that $a, b, c$, and $d$ are positive integers such that

$$
a ! \cdot b ! \cdot c !=d ! \quad \text { and } \quad a+b+c+d=37
$$

compute the product $a b c d$.

Solution: Without loss of generality, assume $a \leq b \leq c<d$. Note that $d$ cannot be prime, as none of $a$ !, $b$ !, or $c$ ! would have it as a factor. If $d=p+1$ for some prime $p$, then $c=p$ and $a ! b !=p+1$. The least possible values of $a ! b$ ! are $1,2,4,6,24,36,48,120,144,240$, so the case where $d=p+1$ is impossible. If $d \geq 21$, then $a+b+c \leq 16$ and it is impossible to find values of $a$ and $b$ such that $a ! \cdot b !=\frac{d !}{c !}$. If $d=16$, either $a ! b !=16$ or $a ! b !=16 \cdot 15$ or $a ! b !=16 \cdot 15 \cdot 14$. Comparing to the list above, the only possible value $a ! b !$ on the list is $16 \cdot 15=240$ and so $(a, b, c, d)=(2,5,14,16)$ and $a b c d=\mathbf{2 2 4 0}$.

Final answer: 2240