Yesterday, Manhattan GMAT posted a 700 level GMAT question on our blog. Today, they have followed up with the answer:
We can rearrange the equation, putting all the x’s on one side and all the y’s on the other side:
2x – x2 = y2 – 2y
Now, list the values of 2n and n2 for the first several nonnegative integers n. In fact, go ahead and compute the differences both ways (both 2n – n2 and n2 – 2n).
| n | 2n | n2 | 2n – n2 | n2 – 2n |
| 0 | 1 | 0 | 1 | –1 |
| 1 | 2 | 1 | 1 | –1 |
| 2 | 4 | 4 | 0 | 0 |
| 3 | 8 | 9 | –1 | 1 |
| 4 | 16 | 16 | 0 | 0 |
| 5 | 32 | 25 | 7 | –7 |
| 6 | 64 | 36 | 28 | –28 |
From this point on, 2n grows much faster than n2, so the differences explode. This means that in order to have a valid equation (2x – x2 = y2 – 2y), we will have to use small values of the integers. We want values in the 2n – n2 column to match values in the n2 – 2n column, and to maximize the value of |x – y|, we want to pick values from different rows—as far apart as possible.
If we pick x = 0 and y = 3 (or vice versa), then we get a valid equation:
20 – 02 = 32 – 23
1 – 0 = 9 – 8
These values of x and y are as far apart as possible, so we get |x – y| = 3.
The correct answer is (D).
