The list of integers $4,4, x, y, 13$ has been arranged from least to greatest. How many different possible ordered pairs $(x, y)$ are there so that the average (mean) of these 5 integers is itself an integer?
(A) 7
(B) 8
(C) 9
(D) 10
(E) 11
The list of integers $4,4, x, y, 13$ has been arranged from least to greatest, and so $4 \leq x$ and $x \leq y$ and $y \leq 13$.

The sum of the 5 integers is $4+4+x+y+13=21+x+y$ and so the average is $\frac{21+x+y}{5}$. Since this average is a whole number, then $21+x+y$ must be divisible by 5 (that is, $21+x+y$ is a multiple of 5).

How small and how large can the sum $21+x+y$ be?

We know that $4 \leq x$ and $x \leq y$, so the smallest that $x+y$ can be is $4+4=8$.

Since $x+y$ is at least 8 , then $21+x+y$ is at least $21+8=29$.

Using the fact that $x \leq y$ and $y \leq 13$, the largest that $x+y$ can be is $13+13=26$.

Since $x+y$ is at most 26 , then $21+x+y$ is at most $21+26=47$.

The multiples of 5 between 29 and 47 are 30,35,40, and 45 .

When $21+x+y=30$, we get $x+y=30-21=9$.

The only ordered pair $(x, y)$ such that $4 \leq x$ and $x \leq y$ and $y \leq 13$, and $x+y=9$ is $(x, y)=(4,5)$.

Continuing in this way, we determine all possible values of $x$ and $y$ that satisfy the given conditions in the table below.

| Value of $21+x+y$ | Value of $x+y$ | Ordered Pairs $(x, y)$ with $4 \leq x$ and $x \leq y$ and $y \leq 13$ |
| :---: | :---: | :---: |
| 30 | $30-21=9$ | $(4,5)$ |
| 35 | $35-21=14$ | $(4,10),(5,9),(6,8),(7,7)$ |
| 40 | $40-21=19$ | $(6,13),(7,12),(8,11),(9,10)$ |
| 45 | $45-21=24$ | $(11,13),(12,12)$ |

The number of ordered pairs $(x, y)$ such that the average of the 5 integers $4,4, x, y, 13$ is itself an integer is 11 .

ANSWER: (E)