Problem 629

Suppose $n=2^{r}5^{s}p_{3}^{a_{3}}p_{4}^{a_{4}}\cdots p_{k}^{a_{k}}$. Since $2n$ has 64 divisors and $5n$ has 60 divisors, $(r+2)(s+1)\left(a_{3}+1\right)\left(a_{4}+1\right)\cdots\left(a_{k}+1\right)=64$ and $(r+1)(s+2)\left(a_{3}+1\right)\left(a_{4}+1\right)\cdots\left(a_{k}+1\right)=60$. The common divisor of 64 and 60 is 1, 2, or 4. Case 1: $\left(a_{3}+1\right)\left(a_{4}+1\right)\cdots\left(a_{k}+1\right)=4$. $n=2^{2}5^{3}p_{3}p_{4}$ or $n=2^{2}5^{3}p_{3}^{3}$. Case 2: $\left(a_{3}+1\right)\left(a_{4}+1\right)\cdots\left(a_{k}+1\right)=2$. No values of $r$ and $s$ work. Case 3: $\left(a_{3}+1\right)\left(a_{4}+1\right)\cdots\left(a_{k}+1\right)=1$. No values of $r$ and $s$ work. Therefore, $n=500p_{3}p_{4}$ or $n=500p_{3}^{3}$. There are 4 possible values of $n$.