Problem (Number Theory):
Compute the largest base-10 integer $\underline{A} \underline{B} \underline{C} \underline{D}$, with $A>0$, such that $\underline{A} \underline{B} \underline{C} \underline{D}=B !+C !+D !$.

Solution:
Let $\underline{A} \underline{B} \underline{C} \underline{D}=N$. Because $7 !=5040$ and $8 !=40,320, N$ must be no greater than $7 !+6 !+6 !=6480$. This value of $N$ does not work, so work through the list of possible sums in decreasing order: $7 !+6 !+5 !, 7 !+6 !+4$ !, etc. The first value that works is $N=5762=7 !+6 !+2 !$.

Let $\underline{A} \underline{B} \underline{C} \underline{D}=N$. Because $7 !=5040$ and $8 !=40,320$, to find the maximal value, first consider values of $N$ that include 7 as a digit. Suppose then that $N=5040+X !+Y$ !. To force a 7 to appear in this sum with maximal $N$, let $X=6$, which yields $N=5040+720+Y !=5760+Y$ !. This value of $N$ has a 7 (and a 6 ), so search for values of $Y$ to find ones that satisfy the conditions of the problem. Only $Y=1$ and $Y=2$ will do, giving 5761 and 5762 . Hence $\mathbf{5 7 6 2}$ is the maximum possible value of $N$.

Final answer: 5762