Question

Let $N$ be a positive integer such that every digit in $N$ is a $1$ or a $2,$ and $N$ is divisible by both $4$ and $9.$ What is the smallest possible value of $N?$

Answer 1

`22212`.

Explanation:

In base-10, for a number to be divisible by
`4`, the last two digits must also be divisible. Among all the two-digit numbers formed with "
`\texttt{1}`" and "
`\texttt{2}`", only
`12` satisfies the requirements as the last two digits of
`N`.

For a number to be divisible by
`9`, the sum of all the digits must also be divisible. For
`N\!\!` to be as small as possible, its digits should add up to the smallest multiple,
`1 * 9 = 9`, instead of larger ones such as
`18`.

Since the last two digits of
`N` is
`\texttt{1}` and
`\texttt{2}`, the other digits of
`N\!` would need to add up to
`9 - 1 - 2 = 6`. To minimize the value of
`N\!\!`, these digits should consist solely of
`\texttt{2}`:
`2 + 2 + 2 = 6`. Therefore, the answer should be
`22212`.