If $x$ and $y$ are positive integers with $x+y=31$, then the largest possible value of $x y$ is
(A) 240
(B) 238
(C) 255
(D) 248
(E) 242
First, we note that the values of $x$ and $y$ cannot be equal since they are integers and $x+y$ is odd.

Next, we look at the case when $x>y$.

We list the fifteen possible pairs of values for $x$ and $y$ and the corresponding values of $x y$ :

| $x$ | $y$ | $x y$ | $x$ | $y$ | $x y$ | $x$ | $y$ | $x y$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 30 | 1 | 30 | 25 | 6 | 150 | 20 | 11 | 220 |
| 29 | 2 | 58 | 24 | 7 | 168 | 19 | 12 | 228 |
| 28 | 3 | 84 | 23 | 8 | 184 | 18 | 13 | 234 |
| 27 | 4 | 108 | 22 | 9 | 198 | 17 | 14 | 238 |
| 26 | 5 | 130 | 21 | 10 | 210 | 16 | 15 | 240 |

Therefore, the largest possible value for $x y$ is 240 .

Note that the largest value occurs when $x$ and $y$ are as close together as possible.

The case of $x<y$ will give us exactly the same result. We can see this by switching the headings of $x$ and $y$ in the table above.

ANSWER: (A)

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