2. The integers $m$ and $n$ satisfy the equation $3^{m} \times n=7!+8!+9$ !. What is the smallest possible value for $n$ ?
SolUTION
560
$$
\begin{array}{l}
7!+8!+9!=7!(1+8+8 \times 9)=7!\times 81= \\
1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 3^{4}=1 \times 2 \times 3 \times 4 \times 5 \times 2 \times 3 \times 7 \times 3^{4}=1 \times 2 \times 4 \times 5 \times 2 \times 7 \times 3^{6} .
\end{array}
$$

The smallest value of $n$ corresponds with $m=6$, and is $1 \times 2 \times 4 \times 5 \times 2 \times 7=560$.