[Number Theory]
Problem: For each positive integer $N$, let $P(N)$ denote the product of the digits of $N$. For example, $P(8)=8$, $P(451)=20$, and $P(2023)=0$. Compute the least positive integer $n$ such that $P(n+23)=P(n)+23$.

Solution: One can verify that no single-digit positive integer $n$ satisfies the conditions of the problem.

If $n$ has two digits, then $n+23$ cannot be a three-digit number; this can be verified by checking the numbers $n \geq 88$, because if $n<88$, then one of the digits of $n+23$ is 0 . Therefore both $n$ and $n+23$ must be two-digit numbers, so the only possible carry for $n+23$ will occur in the tens place. If there is a carry for $n+23$, then $n=\underline{a} \underline{8}$ or $n=\underline{a} \underline{9}$, while $n+23=(a+3) 1$ or $n+23=(a+3) 2$, respectively (the case $n=\underline{a} \underline{7}$ is omitted because then $P(n+23)=0)$. In either case, $P(n+23)<P(n)$ because $a \geq 1$. Otherwise, assume $n=\underline{a} \underline{b}$ and $n+23=(a+2)(b+3)$ is a solution to the given equation, which implies

$$
23=P(n+23)-P(n)=(a+2)(b+3)-a b=3 a+2 b+6 \text {. }
$$

This means $3 a+2 b=17$, which has solutions $(a, b)=(5,1),(3,4)$ as $a, b$ are digits and $b<7$. The two-digit solutions are $n=34$ or $n=51$; thus the least $n$ such that $P(n+23)=P(n)+23$ is $n=34$.

Final answer: 34