Manhattan GMAT Challenge Problem of the Week - 6 June 2011

by Manhattan Prep, Jun 6, 2011

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Question

If two integers are chosen at random out of the set {2, 5, 7, 8}, what is the probability that their product will be of the form [pmath]a^2[/pmath] [pmath]b^2[/pmath], where a and b are both positive integers?

(A) 2/3

(B) 1/2

(C) 1/3

(D) 1/4

(E) 1/6

Answer

Lets first figure out what pairs of integers succeed that is, the product of the two integers is of the desired form [pmath]a^2[/pmath] [pmath]b^2[/pmath].

As a matter of instinct, we should immediately factor [pmath]a^2[/pmath] [pmath]b^2[/pmath] into (a + b)(a b), which is a product. Thus, one way to find the chosen integers is to match the larger one to a + b and the other integer to a b.

Lets say that we pick 2 and 5. Then a + b = 5 and a b = 2. When we solve for a and b, we get a = 3.5 and b = 1.5. Thus, we can't write 10 (the product of 2 and 5) as [pmath]a^2[/pmath] [pmath]b^2[/pmath] in this fashion. Let's try factoring 10 another way into a + b and a b. The only other positive factor pair of 10 is 10 and 1. Solving for a and b, we get a = 5.5 and b = 4.5. So the product of 2 and 5 (i.e., 10) cannot be written as the difference of two perfect squares, [pmath]a^2[/pmath] [pmath]b^2[/pmath], and the pair {2, 5} would be an unsuccessful pair of chosen integers.

Rather than calculate a and b for every pair of integers, lets consider how we might have solved for a in the previous cases. If we chose to factor 10 as 25, we could have added the two equations a + b = 5 and a b = 2, yielding 2a = 2 + 5 = 7. We can see right away that a cannot be an integer, because 7 is an odd number.

Likewise, if we break down 10 as 101, we would get 2a = 10 + 1 = 11, so again, a is not an integer. So when the sum of the two factors is odd, we do not get a possible value of a. If one factor is odd and the other is even, we'll get an odd sum.

All of this means that to get integer values of a and b in the form [pmath]a^2[/pmath] [pmath]b^2[/pmath], we have to be able to break down the product of the chosen integers into at least one factor pair in which both factors are odd or both factors are even.

We need to find out how many of the products can be expressed as either the product of two odd factors or the product of two even factors. All the possible products are 10 (25), 14 (27), 16 (28), 35 (57), 40 (58), and 56 (78).

16 = 2 8. In this case, a = 5 and b = 3.

35 = 5 7. In this case, a = 6 and b = 1.

40 = 4 10. Note that the factors are not the chosen integers 8 and 5 -- we need to pull at least one 2 over from the 8 to make both factors even. In this case, we get a = 7 and b = 3. As an alternative, we could choose 2 and 20 for our factors.

56 = 4 14. a = 9 and b = 5. Again the factors are not the chosen integers 8 and 7 -- we have to pull at least one 2 over from the 8 to make both factors even.

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

The correct answer is A.

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