HMMT-Nov 2018 team, Problem 3

For how many positive integers $n \leq 100$ is it true that $10 n$ has exactly three times as many positive divisors as $n$ has?

Solution: Let $n=2^{a} 5^{b} c$, where $2,5 \nmid c$. Then, the ratio of the number of divisors of $10 n$ to the number of divisors of $n$ is $\frac{a+2}{a+1} \frac{b+2}{b+1}=3$. Solving for $b$, we find that $b=\frac{1-a}{2 a+1}$. This forces $(a, b)=(0,1),(1,0)$. Therefore, the answers are of the form $2 k$ and $5 k$ whenever $\operatorname{gcd}(k, 10)=1$. There are 50 positive numbers of the form $2 k$ and 20 positive numbers of the form $5 k$ less than or equal to 100 . Of those 70 numbers, only $\frac{1}{2} \cdot \frac{4}{5}$ have $k$ relatively prime to 10 , so the answer is $70 \cdot \frac{1}{2} \cdot \frac{4}{5}=28$. $\fbox{28}$.

Answer: 28