If the average (arithmetic mean) of the four numbers 3, 15, 32, and (*N* + 1) is 18, then *N = *

(A) 19

(B) 20

(C) 21

(D) 22

(E) 29

The sum is almost always a more useful piece of information than the average when solving basic stats problems like this. If the average is 18 and there are four numbers, then the sum must be \({4}\times{18} = 72\). So,

\(3 + 15 + 32 + (N + 1) = 72\)

\(51 + N = 72\)

\(N = 21\)

**The correct answer is (C).**