Sajjad1994 wrote:
If two distinct integers x and y are randomly selected from the integers between 1 and 10, inclusive, what is the probability that \(\frac{x^2-y^2}{4}\) is a positive odd integer?
(A) 1/12
(B) 1/14
(C) 1/15
(D) 1/8
(E) 2/15
Solution:
In order for (x^2 - y^2)/4 to be an integer, two integers of the same parity (i.e., both odd or both even) must be selected. To show this, we can let y = x - 2n where n is a positive integer no more than 4 (notice that x and x - 2n have the same parity). Therefore, we have:
x^2 - y^2 = x^2 - (x - 2n)^2 = x^2 - (x^2 - 4nx + 4n^2) = 4nx - 4n^2 = 4n(x - n)
Notice 4n(x - n) is a multiple of 4 and therefore, when it’s divided by 4, the result, n(x - n), is an integer. However, in order for n(x - n) to be a positive odd integer integer, both n and x - n must be odd, and x - n must be positive. Since n is no more than 4, n could be only 1 or 3.
If n = 1, then x = 2, 4, 6, 8, or 10 so that x - n is odd and positive. However, x can’t be 2; otherwise, y = x - 2n would be 0. In other words, we have (x, y) = (4, 2), (6, 4), (8, 6), or (10, 8).
If n = 3,then x = 4, 6, or 8 so that x - n is odd and positive. However, x can’t be 4 or 6; otherwise, y = x - 2n would be negative or 0. In other words, we have only (x, y) = (8, 2) or (10, 4).
We see that there are only 6 ordered pairs of (x, y) such that (x^2 - y^2)/4 is a positive odd integer. Since there are 10P2 = 10 x 9 = 90 ways to choose 2 distinct integers from 1 to 10, inclusive, the probability that (x^2 - y^2)/4 is a positive odd integer is 6/90 = 1/15.
Answer: C
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