For what value of $n$ is $5 \times 8 \times 2 \times n = 7!$?
For example, $5 \times 8 \times 2 \times n = 5 \times 2^3 \times 2 \times n = 5 \times 2^4 \times n$. Similarly, $7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 7 \times 2 \times 3 \times 5 \times 2^2 \times 3 \times 2 = 2^4 \times 3^2 \times 5 \times 7$. Now I can see that both expressions have the same factors of $2^4$ and $5$, but the left-hand side has an extra factor of $n$ and the right-hand side has an extra factor of $3^2 \times 7$. Therefore, to make the equation true, I need $n$ to be equal to $3^2 \times 7 = 9 \times 7 = 63$. 