HMMT-Feb 2023 alg, Problem 5

Suppose $E, I, L, V$ are (not necessarily distinct) nonzero digits in base ten for which \begin{itemize} the four-digit number $\underline{E} \underline{V} \underline{I} \underline{L}$ is divisible by 73 , and the four-digit number $\underline{V} \underline{I} \underline{E}$ is divisible by 74 . \end{itemize} Compute the four-digit number $\underline{L} \underline{I} \underline{V} \underline{E}$.

Solution: Solution: Let $\underline{E}=2 k$ and $\underline{V} \underline{I} \underline{L}=n$. Then $n \equiv-2000 k(\bmod 73)$ and $n \equiv-k / 5(\bmod 37)$, so $n \equiv 1650 k(\bmod 2701)$. We can now exhaustively list the possible cases for $k$ : \begin{itemize}   \item if $k=1$, then $n \equiv 1650$ which is not possible;   \item if $k=2$, then $n \equiv 2 \cdot 1650 \equiv 599$, which gives $E=4$ and $n=599$;   \item if $k=3$, then $n \equiv 599+1650 \equiv 2249$ which is not possible;   \item if $k=4$, then $n \equiv 2249+1650 \equiv 1198$ which is not possible. \end{itemize} Hence, we must have $(E, V, I, L)=(4,5,9,9)$, so $\underline{L} \underline{I} \underline{V} \underline{E}=9954$. $\fbox{9954}$.

Answer: 9954