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w, x, y and z are positive integers. When w is divided by x, the quoti
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13 May 2018, 11:01

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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) x^3 - 3x^2 + 2x = 0 (2) The least common multiple of w and x is 30.

*kudos for all correct solutions

Nice question.

So all of w, x, y, z should be integers greater than 0.

(1) This can be written as x*(x^2 - 3x + 2) = 0 OR x*(x-1)*(x-2) = 0 So x can take 3 values; 0 or 1 or 2. But we are given that x is a positive integer so x cannot be 0. Also if x=1, then when we divide a positive integer w by x (1), the remainder would be 0 (1 is a factor of every positive integer). But the remainder z cannot be 0, thats also given. So we are only left with one value of x, i.e., x=2.

So if x=2, then when we divide any positive integer w by 2, only two remainders are possible: either 0 (if w is even) or 1 (if w is odd). But remainder z cannot be 0 because its a positive integer, so remainder z can only be '1' in this case. Sufficient.

(2) LCM of w and x = 30. We can have w=15, x=6 in which case remainder z will be 3. We can have w=15, x=2 in which case remainder z will be 1. Not sufficient.

Re: w, x, y and z are positive integers. When w is divided by x, the quoti
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13 May 2018, 11:15

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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) x^3 - 3x^2 + 2x = 0 (2) The least common multiple of w and x is 30.

*kudos for all correct solutions

Let me give a try here.

Statement 1 :-

\(x^3-3x^2+2x=0\).

\(x (x^2 -3x+2) = 0\)

x * {x(x-2)-1(x-2)} = 0.

x * (x-1) * (x-2) = 0.

x = 0. is not possible. x = 1 is not correct because. remainder (z) = 0. Hence x = 2.

Any number divided by 2 will yield remainder as ZERO or ONE. So Z has two possibility z= 0 or z=1. But it is given that z is also positive integer. Hence z=1.

Note : Z=1 is possible only if W is odd.

Hence it is sufficient.

Statement 2:-

(2) The least common multiple of w and x is 30. Let's assume value of w & x. w = 15, x= 2 . z=1 .

w=15 , x=6. z = 3.

We are getting two different values of z.

Hence not sufficient.

Ans - A.

Please let me know if I am missing anything. _________________

Re: w, x, y and z are positive integers. When w is divided by x, the quoti
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13 May 2018, 11:41

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amanvermagmat wrote:

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) x^3 - 3x^2 + 2x = 0 (2) The least common multiple of w and x is 30.

*kudos for all correct solutions

Nice question.

So all of w, x, y, z should be integers greater than 0.

(1) This can be written as x*(x^2 - 3x + 2) = 0 OR x*(x-1)*(x-2) = 0 So x can take 3 values; 0 or 1 or 2. But we are given that x is a positive integer so x cannot be 0. Also if x=1, then when we divide a positive integer w by x (1), the remainder would be 0 (1 is a factor of every positive integer). But the remainder z cannot be 0, thats also given. So we are only left with one value of x, i.e., x=2.

So if x=2, then when we divide any positive integer w by 2, only two remainders are possible: either 0 (if w is even) or 1 (if w is odd). But remainder z cannot be 0 because its a positive integer, so remainder z can only be '1' in this case. Sufficient.

(2) LCM of w and x = 30. We can have w=30, x=1 in which case remainder z will be 0. We can have w=15, x=2 in which case remainder z will be 1. Not sufficient.

Re: w, x, y and z are positive integers. When w is divided by x, the quoti
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13 May 2018, 23:21

Mo2men wrote:

amanvermagmat wrote:

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) x^3 - 3x^2 + 2x = 0 (2) The least common multiple of w and x is 30.

*kudos for all correct solutions

Nice question.

So all of w, x, y, z should be integers greater than 0.

(1) This can be written as x*(x^2 - 3x + 2) = 0 OR x*(x-1)*(x-2) = 0 So x can take 3 values; 0 or 1 or 2. But we are given that x is a positive integer so x cannot be 0. Also if x=1, then when we divide a positive integer w by x (1), the remainder would be 0 (1 is a factor of every positive integer). But the remainder z cannot be 0, thats also given. So we are only left with one value of x, i.e., x=2.

So if x=2, then when we divide any positive integer w by 2, only two remainders are possible: either 0 (if w is even) or 1 (if w is odd). But remainder z cannot be 0 because its a positive integer, so remainder z can only be '1' in this case. Sufficient.

(2) LCM of w and x = 30. We can have w=30, x=1 in which case remainder z will be 0. We can have w=15, x=2 in which case remainder z will be 1. Not sufficient.

Re: w, x, y and z are positive integers. When w is divided by x, the quoti
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14 May 2018, 10:47

Top Contributor

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.

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

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