The positive integer $n$ has exactly 8 positive divisors including 1 and $n$. Two of these divisors are 14 and 21 . What is the sum of all 8 positive divisors of $n$ ?
(A) 35
(B) 47
(C) 53
(D) 96
(E) 103

## Part C: Each correct answer is worth 8.
Each of 14 and 21 is a divisor of $n$.

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

Since $21=3 \times 7$, 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 $2 \times 3=6$ is a divisor of $n$.

Since 2, 3 and 7 are divisors of $n$, then their product $2 \times 3 \times 7=42$ 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 $1+2+3+6+7+14+21+42=96$.

ANSWER: (D)