Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

w, x, y and z are positive integers. When w is divided by x, the quoti
[#permalink]

Show Tags

13 May 2018, 11:01

2

1

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
[#permalink]

Show Tags

13 May 2018, 11:15

1

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
[#permalink]

Show Tags

13 May 2018, 11:41

1

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
[#permalink]

Show Tags

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
[#permalink]

Show Tags

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

Answer: A

RELATED VIDEO FROM OUR COURSE

_________________

Brent Hanneson – Founder of gmatprepnow.com

gmatclubot

Re: w, x, y and z are positive integers. When w is divided by x, the quoti &nbs
[#permalink]
14 May 2018, 10:47