A Riddle 2 You

Let us call the integer of interest $N$ .

$N$ 's digits cannot all be $2$ , because $2N$ would consist of all $4$ s. Therefore, at least one of $N$ 's digits must be an integer other than $2$ .

The only one-digit integer that could possibly be $N$ is $2$ , but doubling that would give $4$ , so $N \neq 2$ .

If $N$ were a two-digit integer, $2$ would either be in the tens place or the ones place $\left(N \neq 22\right)$ . In either case, $2N$ will not have two $2$ s, as required. Thus, $N$ cannot be a two-digit integer.

Now let's check the three-digit integers, starting with $2$ in the ones place in an effort to minimize $N$ . We quickly see that $\boxed{N = 112}$ is the minimum $N$ that satisfies our conditions: $2N = 224$ , which has twice as many $2$ s as $N$ .