w, x, y and z are positive integers. When w is divided by x,

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w, x, y and z are positive integers. When w is divided by x, the quotient is y, and the remainder is z. What is the value of z?

1) x³ - 3x² + 2x = 0
2) The least common multiple of w and x is 30.

Answer: A
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Difficulty level: 600-650
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by Brent@GMATPrepNow » Tue May 15, 2018 5:27 am

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Brent@GMATPrepNow wrote:w, x, y and z are positive integers. When w is divided by x, the quotient is y, and the remainder is z. What is the value of z?

1) x³ - 3x² + 2x = 0
2) The least common multiple of w and x is 30.

Answer: A
Source: www.gmatprepnow.com
Difficulty level: 600-650
Target question: What is the value of z?

Given: w, x, y and z are positive integers. When w is divided by x, the quotient is y, and the remainder is z.

Statement 1: x³ - 3x² + 2x = 0
Let's solve this equation for x.
Factor to get: x(x² - 3x + 2) = 0
Factor the quadratic to get: x(x - 1)(x + 2) = 0
So, x = 0, OR x = 1 OR x = 2

Let's examine all 3 cases (x = 0, OR x = 1 OR x = 2):

x = 0: Since we're told that x is a POSITIVE integer, we know that x cannot equal zero.

x = 1: Consider this important rule:
When positive integer N is divided by positive integer D, the remainder R is such that 0 ≤ R < D
For example, if we divide some positive integer by 7, the remainder will be 6, 5, 4, 3, 2, 1, or 0

Likewise, if x = 1, then the remainder (when w is divided by x) must be ZERO.
In other words, if x = 1, then z must equal 0 (according to the above rule)
Since we're told that z is a POSITIVE integer, we know that z cannot equal 0.
So, we can conclude that x cannot equal 1.

By the process of elimination, we know that x = 2
If x = 2 then, according to the above rule, the remainder (z) must equal 0 or 1
However, since we're told that z is a POSITIVE integer, we know that z must equal 1

Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: The least common multiple (LCM) of w and x is 30.
Let's TEST some values of w and x that satisfy statement 2:
Case a: w = 15 and x = 2 (the LCM of 15 and 2 is 30). In this case, w divided by x = 15 divided by 2, in which case the remainder is 1. So, the answer to the target question is z = 1
Case b: w = 2 and x = 15 (the LCM of 15 and 2 is 30). In this case, w divided by x = 2 divided by 15, in which case the remainder is 2. So, the answer to the target question is z = 2
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Answer: A
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