Putnam 2014, Problem B-1

A base $10$ over-expansion of a positive integer $N$ is an expression of the form
$$N = d_k 10^k + d_{k-1} 10^{k-1} + \cdots + d_0 10^0$$
with $d_k \neq 0$ and $d_i \in \{0, 1, 2, \ldots, 10\}$ for all $i$. For instance, the integer $N = 10$ has two base $10$ over-expansions: $10 = 10 \cdot 10^0$ and the usual base $10$ expansion $10 = 1 \cdot 10^1 + 0 \cdot 10^0$. Which positive integers have a unique base $10$ over-expansion?

Solution:
These are integers with no 0s in their base 10 expansion. If the base 10 expansion of \(N\) is \(d_k 10^k + \dots + d_0 10^0\) with a 0, then there exists \(i \leq k-1\) with \(d_i = 0\) and \(d_{i+1} > 0\); replacing \(d_{i+1} 10^{i+1} + d_i 10^i\) by \((d_{i+1} - 1)10^{i+1} + 10 \cdot 10^i\) gives a second base 10 over-expansion. Conversely, if \(N\) has no 0s, its standard form is the unique base 10 over-expansion. By induction on the number of digits: if \(1 \leq N \leq 9\), it is clear. Otherwise, any over-expansion \(N = d_k 10^k + \dots + d_0 10^0\) has \(d_0 \equiv N \pmod{10}\), so \(d_0\) is unique (since \(N\) is not a multiple of 10). Then \((N - d_0)/10\) inherits the over-expansion, unique by induction.