If \(x\) and \(y\) are consecutive positive integers such that \(x < y,\) which of the following statements is true with

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If \(x\) and \(y\) are consecutive positive integers such that \(x < y,\) which of the following statements is true without any exceptions?

I. \((x+1)(y-1) = xy\)
II. \((x+y)^2\) leaves a remainder of \(1\) when divided by \(8.\)
III. The difference between the larger number and the sum of the remainders when \(x\) and \(y\) are divided by each other is \(1.\)

A. I only
B. II only
C. III only
D. I and II only
E. II and III only

Answer: D

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Gmat_mission wrote:
Thu Dec 17, 2020 1:14 pm
If \(x\) and \(y\) are consecutive positive integers such that \(x < y,\) which of the following statements is true without any exceptions?

I. \((x+1)(y-1) = xy\)
II. \((x+y)^2\) leaves a remainder of \(1\) when divided by \(8.\)
III. The difference between the larger number and the sum of the remainders when \(x\) and \(y\) are divided by each other is \(1.\)

A. I only
B. II only
C. III only
D. I and II only
E. II and III only

Answer: D

Solution:

Let’s analyze each statement:

I. (x + 1)(y - 1) = xy

Simplifying, we have:

xy - x + y - 1 = xy

y = x + 1

Since x and y are consecutive positive integers and x < y, then y = x + 1 is true. Therefore, statement I is true.

II. (x + y)^2 leaves a remainder of 1 when divided by 8.

Since x and y are consecutive positive integers, one of them can be written as 2k and the other as 2k + 1 for some positive integer k and therefore, x + y = 4k + 1. Let’s square it:

(x + y)^2 = (4k + 1)^2 = 16k^2 + 8k + 1

Since both 16k^2 and 8k are multiples of 8, we see that (x + y)^2 does leave a remainder of 1 when divided by 8. Statement II is true also.

At this point, by looking at the choices, we see that D must be the correct answer without analyzing statement III.

Answer: D

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