[Combinatorics]
Problem: Let $N$ be a six-digit number formed by an arrangement of the digits $1,2,3,3,4,5$. Compute the smallest value of $N$ that is divisible by 264 .

Solution: Note that $264=3 \cdot 8 \cdot 11$, so we will need to address all these factors. Because the sum of the digits is 18 , it follows that 3 divides $N$, regardless of how we order the digits of $N$. In order for 8 to divide $N$, we need $N$ to end in $\underline{O} 12, \underline{O} 52, \underline{E} 32$, or $\underline{E} 24$, where $O$ and $E$ denote odd and even digits. Now write $N=\underline{U} \underline{V} \underline{W} \underline{X} \underline{Y} \underline{Z}$. Note that $N$ is divisible by 11 if and only if $(U+W+Y)-(V+X+Z)$ is divisible by 11. Because the sum of the three largest digits is only 12 , we must have $U+W+Y=V+X+Z=9$.

Because $Z$ must be even, this implies that $V, X, Z$ are $2,3,4$ (in some order). This means $Y \neq 2$, and so we must have $Z \neq 4 \Rightarrow Z=2$. Of the three remaining possibilities, $\underline{E} 32$ gives the smallest solution, 135432.

Final answer: 135432