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BrentGMATPrepNow
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w, x, y and z are positive integers. When w is divided by x, the quoti 13 May 2018, 06:47
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

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95% (hard)

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24% (02:12) correct 76% (02:40) wrong based on 300 sessions

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

*kudos for all correct solutions
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amanvermagmat
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w, x, y and z are positive integers. When w is divided by x, the quoti 13 May 2018, 11:01
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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) x^3 - 3x^2 + 2x = 0
(2) The least common multiple of w and x is 30.

*kudos for all correct solutions
Show more

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.

Hence A answer
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w, x, y and z are positive integers. When w is divided by x, the quoti 13 May 2018, 11:15
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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) x^3 - 3x^2 + 2x = 0
(2) The least common multiple of w and x is 30.

*kudos for all correct solutions
Show more


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

Hence A answer
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amanvermagmat

In statement 1, you rejected z as z should positive but again in statement 2 you consider that value of z may be zero. You should discard it too.
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w, x, y and z are positive integers. When w is divided by x, the quoti 13 May 2018, 11:52
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GMATPrepNow
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
Show more

Given: w, x, y and z are positive integers

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

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

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

Lets's examine the 3 possibilities:

If x = 0.........x is divisor.......... so x can't be 0, otherwise it would be indefinite.......Discard

If x = 1........z would be 0..........however z is positive.......Discard

If x = 2........z would take either 0 or 1 but z is positive so we left with 1.

Sufficient

(2) The least common multiple of w and x is 30.

Let W = 15, X = 2.........z =1

Let W = 6, X = 5...........z =1

Let W = 15, X = 6.........z =3

Insufficient

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

Hence A answer

amanvermagmat

In statement 1, you rejected z as z should positive but again in statement 2 you consider that value of z may be zero. You should discard it too.
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Hello

Yes I have done that and edited my post. Thank you.
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w, x, y and z are positive integers. When w is divided by x, the quoti Updated on: 08 Aug 2022, 11:31
Originally posted by BrentGMATPrepNow on 14 May 2018, 10:47.
Last edited by BrentGMATPrepNow on 08 Aug 2022, 11:31, edited 3 times in total.
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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.
Show more

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 = 10 and x = 6 (the LCM of 10 and 6 is 30). In this case, w divided by x = 10 divided by 6, in which case the remainder is 4. So, the answer to the target question is z = 4
* case b is courtesy of u/djosh7 from Reddit. Thanks djosh7

Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Answer: A

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