Japan Mathematical Olympiad First Stage, 2019

235

Let $n$ be a three-digit good number, then we can write $n = 100a + 10b + c$, where $a, b, c$ are 1-digit primes. If $n^2$ is a good number with 5 digits, we must have $n^2 < 10^5$, from which it follows that we have $n < 320$. Since $10b + c \ge 22$, we conclude that $a = 2$ must hold. When $c = 2, 3, 5, 7$, the lowest digit of $n^2$ is 4, 9, 5, 9, respectively. If $n^2$ is also a good number, the lowest digit of $n^2$ must be 5, which, in turn, implies that $c = 5$ must hold.

Since $a = 2$ and $c = 5$, we conclude that the possibilities for the choice of $n$ is 225, 235, 255, 275. But, then $n^2 = 50625, 55225, 65025, 75625$, among which $55225 = 235^2$ is the only good number. Therefore, we conclude that $n = 235$ is the desired answer.