AAPL wrote:Manhattan Prep
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 a^2 - b^2, where a and b are both positive integers?
A. 2/3
B. 1/2
C. 1/3
D. 1/4
E. 1/6
OA A
Since 4C2 = 6, we only have 6 ways to choose 2 numbers from the 4 numbers in the set. We can just list the pairs of number and their products:
2 x 5 = 10
2 x 7 = 14
2 x 8 = 16
5 x 7 = 35
5 x 8 = 40
7 x 8 = 56
Now we need to determine which of these products are in the form of a^2 - b^2 (difference of squares), where a and b are positive integers. Let's list the first 20 positive perfect squares:
1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, ...
We see that the difference of a pair of consecutive perfect squares is odd, and we see that 324 - 289 = 18^2 - 17^2 = 35, which is one of the products and the only odd product.
Now, since the remaining 5 products are even, they must be a difference of two non-adjacent perfect squares. Let's analyze each of these 5 products:
10: Since 9 - 1 = 8 and 16 - 4 = 12, we see that there is no way we can get a difference of 10.
14: Since 16 - 4 = 12 and 25 - 9 = 16, we see that there is no way we can get a difference of 14.
16: Since 25 - 9 = 16, we see that 16 = 5^2 - 3^2.
40: Since 121 - 81 = 16, we see that 40 = 11^2 - 9^2.
56: Since 225 - 169 = 56, we see that 56 = 15^2 - 13^2.
Since 4 of the 6 products can be written as a^2 - b^2, the probability is 4/6 = 2/3.
Alternate solution:
We can use the following facts:
If a and b are positive integers:
1) Any odd number greater than or equal to 3 can be written in the form of a^2 - b^2.
2) Any multiples of 4 greater than or equal to 8 can be written in the form of a^2 - b^2.
3) Any other positive integers (i.e., 1, 4, and even numbers that are not multiples of 4) can't be written in the form of a^2 - b^2.
Therefore, from the 6 products: 10, 14, 16, 35, 40, and 56, we see that the 1st fact applies to 35, the second fact applies to 16, 40, and 56, and the 3rd fact applies to 10 and 14. Facts 1) and 2) gives us 4 numbers that work, and fact 3) gives us 2 numbers that don't work. Thus, 4 of the 6 products can be written in the form of a^2 - b^2, resulting in a probability of 4/6 = 2/3.
Answer: A
