South African Mathematics Olympiad

The required numbers are of the form $3^a4^b5^c6^d7^e$, where
$$0 \le a \le 1, \quad 0 \le b \le 2, \quad 0 \le c \le 2, \quad 0 \le d \le 1, \quad 0 \le e \le 3,$$
and $a+b+c+d+e \ge 2$. This gives two possible values of $a$, three of $b$, three of $c$, two of $d$, and four of $e$, making a total of $2 \times 3 \times 3 \times 2 \times 4 = 144$ possible numbers if we ignore the last restriction. To satisfy it, we must exclude one possibility with $a+b+c+d+e = 0$ and five possibilities with $a+b+c+d+e = 1$, so the final total is $144 - 1 - 5 = 138$.