Multiple of 45 using only the digits 0 and 5

Statement

What is the smallest positive multiple of $45$ that can be written using only the digits $0$ and $5$?

Solution

Since $45 = 9\cdot 5$ with $9$ and $5$ coprime, the number must be divisible by $9$ and by $5$.

• By $5$: it must end in $0$ or $5$: always possible with the available digits.
• By $9$: the digit sum must be divisible by $9$. Using only $0$ and $5$, the sum is $5k$, where $k$ is the number of $5$-digits. We need $9 \mid 5k$, i.e. $9 \mid k$: at least nine $5$-digits are required.

The smallest such number uses nine $5$-digits (inserting $0$s would only make it longer): $\boxed{555\,555\,555}.$