jmc

Divisibility by $33$ requires that a number be divisible by both $11$ and $3$. If a five-digit number is divisible by $11$, the difference between the sum of the units, hundreds and ten-thousands digits and the sum of the tens and thousands digits must be divisible by $11$. Thus $(7+9+3)-(n+3)=16-n$ must be divisible by $11$. The only digit which can replace $n$ for the number to be divisible by $11$, then, is $n=5$. Furthermore, if a number is $7+5+9+3+3=27$, so the number is divisible by $3$. Hence, $n=\boxed{5}$.