HMMT February

Solution:

Let $\underline{E} = 2k$ and $\underline{V}\,\underline{I}\,\underline{L} = n$. Then $n \equiv -2000k \pmod{73}$ and $n \equiv -k/5 \pmod{37}$, so $n \equiv 1650k \pmod{2701}$. We can now exhaustively list the possible cases for $k$:
- if $k=1$, then $n \equiv 1650$ which is not possible;
- if $k=2$, then $n \equiv 2 \cdot 1650 \equiv 599$, which gives $E=4$ and $n=599$;
- if $k=3$, then $n \equiv 599+1650 \equiv 2249$ which is not possible;
- if $k=4$, then $n \equiv 2249+1650 \equiv 1198$ which is not possible.

Hence, we must have $(E, V, I, L) = (4, 5, 9, 9)$, so $\underline{L}\,\underline{I}\,\underline{V}\,\underline{E} = 9954$.