Putnam

These are the integers with no $0$'s in their usual base $10$ expansion. If the usual base $10$ expansion of $N$ is $d_k 10^k + \cdots + d_0 10^0$ and one of the digits is $0$, then there exists an $i \leq k-1$ such that $d_i = 0$ and $d_{i+1} > 0$; then we can replace $d_{i+1} 10^{i+1} + (0) 10^i$ by $(d_{i+1}-1) 10^{i+1} + (10) 10^i$ to obtain a second base $10$ over-expansion. We claim conversely that if $N$ has no $0$'s in its usual base $10$ expansion, then this standard form is the unique base $10$ over-expansion for $N$. This holds by induction on the number of digits of $N$: if $1\leq N\leq 9$, then the result is clear. Otherwise, any base $10$ over-expansion $N = d_k 10^k + \cdots + d_1 10 + d_0 10^0$ must have $d_0 \equiv N \pmod{10}$, which uniquely determines $d_0$ since $N$ is not a multiple of $10$; then $(N-d_0)/10$ inherits the base $10$ over-expansion $d_k 10^{k-1} + \cdots + d_1 10^0$, which must be unique by the induction hypothesis.