South African Mathematics Olympiad

The required numbers are of the form $3^a{4}^b{5}^c{6}^d{7}^e$, where $$0\le a\le 1,0\le b\le 2,0\le c\le 2,0\le d\le 1,0\le e\le 3,$$ 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 $144-1-5=138$.