[Number Theory]
Problem: Compute the base $b$ for which $253_{b} \cdot 341_{b}=\underline{7} \underline{4} \underline{X} \underline{Y} \underline{Z}_{b}$, for some base- $b$ digits $X, Y, Z$.

Solution: Write $253_{b} \cdot 341_{b}=\left(2 b^{2}+5 b+3\right)\left(3 b^{2}+4 b+1\right)=6 b^{4}+23 b^{3}+31 b^{2}+17 b+3$. Compare the coefficients in this polynomial to the digits in the numeral $\underline{7} \underline{4} \underline{X} \underline{Y} \underline{Z}$. In the polynomial, the coefficient of $b^{4}$ is 6 , so there must be a carry from the $b^{3}$ place to get the $7 b^{4}$ in the numeral. After the carry, there should be no more than 4 left for the coefficient of $b^{3}$ as only one $b$ is carried. Therefore $23-b \leq 4$ or $b \geq 19$. By comparing digits, note that $Z=3$. Then

$$
\begin{aligned}
6 b^{4}+23 b^{3}+31 b^{2}+17 b & =\underline{7} \underline{4} \underline{X} \underline{Y} \underline{0} \\
& =7 b^{4}+4 b^{3}+X \cdot b^{2}+Y \cdot b
\end{aligned}
$$

Because $b>0$, this equation can be simplified to

$$
b^{3}+X \cdot b+Y=19 b^{2}+31 b+17
$$

Thus $Y=17$ and $b^{2}+X=19 b+31$, from which $b(b-19)=31-X$. The expression on the left side is positive (because $b>19$ ) and the expression on the right side is at most 31 (because $X>0$ ), so the only possible solution is $b=20, X=11$. The answer is 20 .

Final answer: 20