If $1998=p^{s} q^{t} r^{u}$, where $p, q$ and $r$ are prime numbers, what is the value of $p+q+r$ ?
(A) 222
(B) 48
(C) 42
(D) 66
(E) 122
If $1998=p^{s} q^{t} r^{u}$, where $p, q$ and $r$ are prime numbers, what is the value of $p+q+r$ ?
(A) 222
(B) 48
(C) 42
(D) 66
(E) 122

## Solution

The prime factorization of 1998 is $2 \times 3^{3} \times 37$. Thus, $p, q$, and $r$ have values 2,3 , and 37 (in any order), and $p+q+r=42$.