Problem 595

Each of 14 and 21 is a divisor of $n$.

Since $14=2\times7$, then each of 2 and 7 is also a divisor of $n$.

Since $21=3\times7$, then 3 is a divisor of $n$ (as is 7 which we already noted).

So far, the positive divisors of $n$ are: $1,2,3,7,14$, and 21.

Since 2 and 3 are divisors of $n$, then their product $2\times3=6$ is a divisor of $n$.

Since $2,3$ and 7 are divisors of $n$, then their product $2\times3\times7=42$ is a divisor of $n$.

The positive divisors of $n$ are: $1,2,3,6,7,14,21$, and 42.

We are given that $n$ has exactly 8 positive divisors including 1 and $n$, and so we have found them all, and thus $n=42$.

The sum of these 8 positive divisors is $1+2+3+6+7+14+21+42=96$.