A set of five different positive integers has a mean (average) of 20 and a median of 18. What is the greatest possible integer in the set?
(A) 60
(B) 26
(C) 46
(D) 12
(E) 61
In an ordered list of five integers, the median is the number in the middle or third position. Thus if we let the set of integers be $a, b, c, d, e$, ordered from smallest to largest, then $c=18$. Since the average is fixed (at 20), $e$ (the largest number in the set) is largest when $a, b$ and $d$ are as small as possible.

Since the numbers in the set are different positive integers, the smallest that $a$ can be is 1 and the smallest that $b$ can be is 2 .

Our set of integers is now $1,2,18, d, e$.

Again, we want $d$ to be as small as possible, but it must be larger than the median 18 .

Therefore, $d=19$.

Since the average of the 5 integers is 20 , then the sum of the five integers is $5 \times 20=100$.

Thus, $1+2+18+19+e=100$ or $40+e=100$, and so $e=60$.

The greatest possible integer in the set is 60 .

Answer: (A)