2018/19 AHSMC Part 1 Problem 14

For the first $100$ positive integers the sequence “$31$” occurs twice, once from $13$ to $14$ and again as $31$. Now, let $x$ be a three digit number and suppose “$31$” appears in $x(x+1)$ with the “$3$” appearing in either the ones, tens or hundreds place of $x$. If $x$ ends in $3$, then $x+1$ begins with $1$, which implies $x = 1a3$. Each of $a \in \{0,1,\ldots,9\}$ is possible. If $x$ has tens digit $3$, then we must have $x = a31$. Each of $a \in \{1,2,\ldots,9\}$ is possible. Finally, if $x$ begins with $3$, then we must have $x = 31a$. Each of $a \in \{0,1,\ldots,9\}$ is possible. Thus, the total number of occurrences of “$31$” is $2 + 10 + 9 + 10 = 31$. The answer is (E).