## Blog

### The Quest for 700: Weekly GMAT Challenge (Answer)

Yesterday, Manhattan GMAT posted a GMAT question on our blog. Today, they have followed up with the answer:

Let’s first figure out what pairs of integers “succeed” – that is, the product of the two integers is of the desired form a2b2. As a matter of instinct, we should immediately factor a2b2 into (a + b)(ab), which is a product. Thus, we need one of the two chosen integers (specifically, the larger one) to match a + b, while the other integer matches ab.

Let’s say that we pick 2 and 5. Then a + b = 5 and ab = 2. When we solve for a and b, we get a = 3.5 and b = 1.5. So the product of 2 and 5 (i.e., 10) cannot be written as the difference of two perfect squares, a2b2, and the pair {2, 5} would be an unsuccessful pair of chosen integers.

Rather than calculate a and b for every pair of integers, let’s consider how we might have solved for a in the previous case. We could have added the two equations, yielding 2a = 2 + 5 = 7. We can see right away that a cannot be an integer, because 7 is an odd number. In fact, whenever the sum of the two chosen integers is odd, we will not get integer values of a and b. On the other hand, when the sum of the two chosen integers is even, we will get integer values of a and b.

We need to find out how many of the products can be expressed as either the product of two odd or two even factors. All the possible products are 10 (2×5), 14 (2×7), 16 (2×8), 35 (5×7), 40 (5×8), and 56 (7×8).

16 = 2 × 8. a = 5 and b = 3
35 = 5 × 7. a = 6 and b = 1
40 = 4 × 10. a = 7 and b = 3
56 = 4 × 14. a = 9 and b = 5

There are 6 possible products, and 4 of them can be expressed as the product (a + b)(ab).
Thus, the probability we want is 4/6 or 2/3.

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