South African Mathematics Olympiad

Placing a $3$ at both ends of the number increases it by $3372$, a four-digit number. Hence the original number is a two-digit number. Suppose the original number is $x$. Then the number formed by placing a $3$ at both ends is equal to $3003+10x$. This is $3372$ more than $x$, that is, $$3003+10x-x=3372⟹9x=369⟹x=41.$$