2020/21 AHSMC Part 1 Problem 15

If $x = 10a + b$ then $x^3 = 1000a^3 + 300a^2b + 30ab^2 + b^3$. The remainder when $x^3$ is divided by $100$ is equal to the cube $b^3$ of the units digit of $x$ if and only if $100$ is a divisor of $30ab^2$ and $b^3 < 100$, which means that $10$ is a divisor of $ab^2$ and $b \le 4$. This happens when $a \in \{1,2,3,4,5,6,7,8,9\}$ and $b = 0$, or if $a = 5$ and $b \in \{2,4\}$. Therefore there are $11$ integers in total. The answer is (B).