If two distinct integers x and y are randomly selected from the intege

Question

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

TestPrepUnlimited · Accepted Answer

\frac{x^2-y^2}{4} = \frac{(x + y)(x - y) }{ 4} .

We need (x + y)(x - y) to be divisible by 4 first of all, in order to do that we need both x and y to be both even or both odd. 1 odd 1 even will result in Odd*Odd.

However, if we have both x and y odd, one of x + y or x - y will be a multiple of 4 guaranteed. This is because while they are both even, their difference is 2y which is not a multiple of 4. If x - y is not a multiple of 4 for example, adding 2y will make x + y a multiple of 4 and vice versa. Then we have a multiple of 4 times an even number. After dividing by 4 the result will be even, thus we cannot take any x odd y odd options.

Therefore the only possible options come from x even and y even.

Next, in order to have an odd result after dividing by 4, we need to make sure (x + y) is not a multiple of 4, this guarantees (x - y) is also not a multiple of 4 because 2y is a multiple of 4 (similar logic with above).

Thus picking out even options while avoiding x+y being a multiple of 4, and x > y to guarantee a positive result, we have (4, 2), (6, 4), (8, 2), (8, 6), (10, 4), (10, 8).

There are 10 * 9 random selections since once you choose one of the 10 options for x there are only 9 options left for y. The answer should be 6/90 = 1/15

Ans: C

ScottTargetTestPrep · Answer

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

Hellohello007 · Answer

Hii,

Why havent we included the case with both odd?

Bunuel · Answer

If both x and y are odd, every odd square leaves remainder 1 when divided by 8. Therefore, x^2 - y^2 is divisible by 8, and after dividing by 4 the result is even, not odd.

Hellohello007 · Answer

and how does that work can you also explain the working math please?

Bunuel · Answer

Take any odd integer as 2k + 1.

(2k + 1)^2 = 4k^2 + 4k + 1 = 4k(k + 1) + 1.

Since k and k + 1 are consecutive integers, one is even, so 4k(k + 1) is divisible by 8. Therefore, every odd square leaves remainder 1 when divided by 8.

So if both x and y are odd, both x^2 and y^2 leave remainder 1 when divided by 8. Hence x^2 - y^2 is divisible by 8, and (x^2 - y^2)/4 is even, not odd.

For example, if x = 5 and y = 3, then (25 - 9)/4 = 16/4 = 4, which is even.

If two distinct integers x and y are randomly selected from the intege

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If two distinct integers x and y are randomly selected from the intege

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