When in $b-$base representation of a number all of $0, 1, 2, ..., b-1$ exist? If $a,b \ge 2$ are given, prove that there is a positive integer $m$ such that $ a \mid m$ and in $b-$base representation of $m$ all of $0, 1, 2, ..., b-1$ exist.
My teacher gave this to me, but I think his statement was wrong. It was like this: 
For $a,b \ge 2$, prove that there is a positive integer $m$ such that $ a \mid m$ and in $b-$base representation of $\color{red}a$ all of $0, 1, 2, ..., b-1$ exist.
Mine makes more sense, Isn't it? 

Pick $r$ with $b^r>a$. Let $A$ be the number that, in base $b$, uses all digits. Then one of the numbers $b^rA,\ldots, b^rA+a-1$ will do.
