2024 SSMO Team Round Problems/Problem 3

Problem

Consider positive integers \(N\) such that when \(N\)'s units digit and leading nonzero digit are removed, what remains is a two-digit perfect square. The average of all \(N\) can be expressed as $\frac{m}{n},$ for relatively prime positive integers $m$ and $n.$ Find $m+n.$

Solution

Note that the two-digit perfect square obtained from removing the leading and units digit has to be in the set $\{16,25,\dots,81\}.$ Taking the expected value of $N,$ we have \begin{align*} \mathbb{E}(N) &= \mathbb{E}(i\in\{1,2,\dots,9\})\cdot1000+\left(\frac{4^2+5^2+\dots+9^2}{6}\right)\cdot10+\mathbb{E}(i\in\{0,1,2\dots,9\})\\ &=5\cdot1000+\left(\frac{271}{6}\right)\cdot10+\frac92 = \frac{32737}{6}\implies 32737+6 = \boxed{32733}. \end{align*}

~SMO_Team