There is the possibility of using only $3$s giving one possible number $333333$.
Let's suppose a second digit is used, say $x$. After the initial digit $3$, there are five positions into which we can put either $3$ or $x$. So there are two choices in each of these five positions and so $2^5 = 32$ possible choices - except that one such choice would be five $3$s. So we get $31$ choices.
There are nine possible values for $x$, namely $0$, $1$, $2$, $4$, $5$, $6$, $7$, $8$, $9$.
So this gives $9\times 31=279$ numbers.