For the positive integers x, a, and b, when x is divided by a, the remainder is b. And, when x is divided by b, the remainder is a – 2. Which of the following statements must be true?

A. a is even. B. x + b is divisible by a. C. x – 1 is divisible by a. D. b = a – 1 E. a + 2 = b + 1

(D) The easiest way to answer this question is by picking numbers. It may take a bit of trial and error to find numbers that satisfy the question stem, but when you do, you can eliminate the wrong answer choices.

Let’s plug in the following values for x and a. Let x = 5 and a = 3. “When x is divided by a, the remainder is b” implies that b is less than a. When we divide 5 by 3, we get 1 with a remainder of 2, so b = 2. We must make sure that these values make the next statement correct as well. When we divide 5 by 2, we get 2 with a remainder of 1, which is what we wanted for a remainder. Therefore, these numbers will help us eliminate answer choices.

Since the question asks which of the choices must be true, we can eliminate any choice that is false for any set of acceptable numbers.

Let’s try the choices with x = 5, a = 3, and b = 2: Choice (A): 3 is even. This is FALSE. Choice (B): 5 + 2 is divisible by 3. This is FALSE. Choice (C): 5 – 1 is divisible by 3. This is FALSE. Choice (D): 2 = 3 – 1. This is TRUE. Choice (E): 3 + 2 = 2 + 1. This is FALSE.

Since Choice (D) is the only choice that survived, it must be the correct answer.

There is an alternative algebraic way of handling it. When x is divided by a, the remainder is b. Therefore, b must be less than a.

When x is divided by b, the remainder is a – 2. Therefore, a - 2 must be less than b.

Thus, a > b > a – 2 ...or, b = a – 1. The correct answer is D. ---------- Would like to know a better way than plugging examples. Because there might be a possibility that the examples I plug in may/may not suit the situation and the answer we are looking at.

Would like to know a better way than plugging examples. Because there might be a possibility that the examples I plug in may/may not suit the situation and the answer we are looking at.

First of all, note that there is the algebraic solution provided in the last part of the explanation.

Secondly, plugging in few easy numbers is a good method that will help to eliminate wrong answers and/or understand a problem better. You may use it if you don't see an algebraic solution. If you eliminate all answer choices except one, then that one is the right answer.

When you plug in numbers, do not pick them all at once, but replace variables one-by-one, trying to fit the new value with the already chosen ones.

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