It is currently 16 Jan 2018, 11:34

### GMAT Club Daily Prep

#### 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

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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# When the positive integer x divided by the positive integer

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 43296

Kudos [?]: 139200 [4], given: 12779

When the positive integer x divided by the positive integer [#permalink]

### Show Tags

17 Jun 2014, 01:58
4
KUDOS
Expert's post
29
This post was
BOOKMARKED
00:00

Difficulty:

85% (hard)

Question Stats:

55% (01:44) correct 45% (01:46) wrong based on 415 sessions

### HideShow timer Statistics

When the positive integer $$x$$ is divided by the positive integer $$y$$, the quotient is 3 and the remainder is $$z$$. When $$z$$ is divided by $$y$$, the remainder is 2. Which of the following could be the value of $$x$$?

I. 5
II. 8
III. 32

A. I only
B. II only
C. III only
D. I and II only
E. I, II and III

Kudos for a correct solution.

[Reveal] Spoiler: OA

_________________

Kudos [?]: 139200 [4], given: 12779

Math Expert
Joined: 02 Sep 2009
Posts: 43296

Kudos [?]: 139200 [4], given: 12779

When the positive integer x divided by the positive integer [#permalink]

### Show Tags

17 Jun 2014, 01:59
4
KUDOS
Expert's post
15
This post was
BOOKMARKED
SOLUTION

When the positive integer is $$x$$ divided by the positive integer $$y$$, the quotient is 3 and the remainder is $$z$$. When $$z$$ is divided by $$y$$, the remainder is 2. Which of the following could be the value of $$x$$?

I. 5
II. 8
III. 32

A. I only
B. II only
C. III only
D. I and II only
E. I, II and III

When $$x$$ is divided by $$y$$, the quotient is 3 and the remainder is $$z$$: $$x=3y+z$$, where $$0\leq{z}<y$$ (the remainder must be less than the divisor).

When $$z$$ is divided by $$y$$, the remainder is 2: when divisor (y in our case) is more than dividend (z in our case), then the reminder equals to the dividend (for example, 2 divided by 5 gives the remainder of 2). Therefore, $$z=2$$ and $$2<y$$.

So, we have that $$x=3y+2$$ and $$2<y$$. This implies that the least value of $$x$$ is $$x=3*3+2=11$$: $$x$$ cannot be 5 or 8.

Could $$x$$ be 32? Yes. If $$y=10$$, then $$x=3*10+2=32$$.

Try NEW remainders DS question.
_________________

Kudos [?]: 139200 [4], given: 12779

Intern
Joined: 28 May 2014
Posts: 16

Kudos [?]: 2 [1], given: 0

Re: When the positive integer x divided by the positive integer [#permalink]

### Show Tags

17 Jun 2014, 02:21
1
KUDOS
My solution might be awful. But I did it the following way:

X/Y = 3+Z/Y.

Z/Y=2/Y.

Then I just plugged in 5 and tried adjusting Y.

There is no way I could get 5/Y= 3+2/Y as Y = 2 gives me 2+1/Y, and 1 would give me 5+0/Y.
Hence I is not possible.

Then 8 wouldn't work either.
Set Y = 2, gives X/Y=4+0/Y, no remainder.
Set Y = 3, gives X/Y=2+2/Y (The remainder is correct here, but we wanted 3+2/Y.
Hence II is not a possible value of X.

Lastly, X = 32.
I directly saw that setting Y = 10 would give me 3+2/Y. I didn't need to try any other options.

Therefore, the answer is C. Only III is a possible value.

Kudos [?]: 2 [1], given: 0

Intern
Joined: 29 May 2013
Posts: 3

Kudos [?]: [0], given: 0

Schools: Booth '17
Re: When the positive integer x divided by the positive integer [#permalink]

### Show Tags

17 Jun 2014, 03:11
x=3y+z---(1)
z=ky+2----(2)
Combining 1 & 2, x=(3+k)y + 2
Already, equation 1 tells us that the remainder when x is divided by y is z.
So, applying the above result to equation 2 would give z=2
Thus x=3y+2
In other words, x when divided by 3 leaves remainder 2. Thus 5,8,32 all satisfy this criteria. Hence the answer is E.

Kudos [?]: [0], given: 0

Senior Manager
Joined: 13 Jun 2013
Posts: 278

Kudos [?]: 499 [1], given: 13

Re: When the positive integer x divided by the positive integer [#permalink]

### Show Tags

17 Jun 2014, 07:50
1
KUDOS
Bunuel wrote:

When the positive integer $$x$$ divided by the positive integer $$y$$, the quotient is 3 and the remainder is $$z$$. When $$z$$ is divided by $$y$$, the remainder is 2. Which of the following could be the value of $$x$$?

I. 5
II. 8
III. 32

A. I only
B. II only
C. III only
D. I and II only
E. I, II and III

Kudos for a correct solution.

when x is divided by y, remainder will always be less than the denominator y.

also, if numerator is less than denominator then remainder is equal to the numerator.

now coming to the question x= 3y+ z ----------------1)
using the above property we can conclude that x>y>z --------------------2)

also when z is divided by y remainder is 2. now since z is less than y therefore z must be equal to 2.

therefore by substituting the value of z=2 in the equation 1 we have

x=3y+2

also from 2 we have y>2

thus values x=5 and x= 8 are eliminated and thus x=32 is the only possible answer.

hence answer should be C

Kudos [?]: 499 [1], given: 13

Intern
Joined: 19 Jan 2008
Posts: 19

Kudos [?]: 6 [2], given: 4

Re: When the positive integer x divided by the positive integer [#permalink]

### Show Tags

17 Jun 2014, 10:27
2
KUDOS
We have got 2 equations :-
Eqn1 - x=3y+z
Eqn2 - z=Qy+2(where Q is the quotient).

32 is the only option that can be written in accordance with equation 1 and leaves remainder of 2 when divided by 10.Hence,answer is C.

Kudos [?]: 6 [2], given: 4

Director
Joined: 25 Apr 2012
Posts: 721

Kudos [?]: 885 [0], given: 724

Location: India
GPA: 3.21
WE: Business Development (Other)
Re: When the positive integer x divided by the positive integer [#permalink]

### Show Tags

17 Jun 2014, 20:42
Bunuel wrote:
When the positive integer $$x$$ divided by the positive integer $$y$$, the quotient is 3 and the remainder is $$z$$. When $$z$$ is divided by $$y$$, the remainder is 2. Which of the following could be the value of $$x$$?

I. 5
II. 8
III. 32

A. I only
B. II only
C. III only
D. I and II only
E. I, II and III

Kudos for a correct solution.

When the positive integer $$x$$ divided by the positive integer $$y$$, the quotient is 3 and the remainder is $$z$$

Can be written as $$x= 3y+z$$ or $$z= x-3y$$

When $$z$$ is divided by $$y$$, the remainder is 2

So we have $$z= by+2$$

or by+2=x-3y or
x = by+3y+2 = 3y+z ---------> by+2=z=x

Now if by= even, then z is even and hence x is even.
but if by is odd then z is odd and thus x is odd

Since we need to find what could be the possible value...then any of three options could be the value...So Ans E
_________________

“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”

Kudos [?]: 885 [0], given: 724

Intern
Joined: 13 May 2014
Posts: 2

Kudos [?]: 1 [1], given: 3

Re: When the positive integer x divided by the positive integer [#permalink]

### Show Tags

18 Jun 2014, 00:27
1
KUDOS
When the positive integer $$x$$ divided by the positive integer $$y$$, the quotient is 3 and the remainder is $$z$$. When $$z$$ is divided by $$y$$, the remainder is 2. Which of the following could be the value of $$x$$?

I. 5
II. 8
III. 32

A. I only
B. II only
C. III only
D. I and II only
E. I, II and III

x = 3y+z. (Now y can't take the value zero as z/y can't be defined)

when x = 5
y can only be 1 and thus z = 2 and the remainder is 0 for z/y

when x = 8
y = 1 and z = 5 and the remainder is 0 for z/y

y = 2 and z = 2 and the remainder is 0 for z/y

thus, only option C is applicable.

when x = 32
y can take 3 or 10 and z will be 23 or 2 respectively which gives the remainder as 2 for z/y.

Kudos [?]: 1 [1], given: 3

Manager
Joined: 04 Sep 2012
Posts: 100

Kudos [?]: 50 [1], given: 504

Location: Philippines
Concentration: Marketing, Entrepreneurship
Schools: Ross (Michigan) - Class of 2017
GMAT 1: 620 Q48 V27
GMAT 2: 660 Q47 V34
GMAT 3: 700 Q47 V38
GPA: 3.25
WE: Sales (Manufacturing)
Re: When the positive integer x divided by the positive integer [#permalink]

### Show Tags

18 Jun 2014, 01:49
1
KUDOS
I did number picking on this one.

$$x/y=3y+z/y$$ ----(1)
$$z/y=by+2/y$$ ----(2)

Both $$x=5$$ & $$x=8$$ are too small to have a quotient of $$3$$ and still have $$y$$ as an integer.

Only $$x=32$$ is big enough to have $$y$$ as a positive integer with a quotient of $$3$$.

With this, only $$y=10$$ can complete the solution.

for $$x/y$$ --> $$32/10=3(10)+2/10$$
for $$z/y$$ --> $$2/10=2$$

Therefore, only $$x=32$$ can be the value of $$x$$ given the restrictions in the question.

Kudos [?]: 50 [1], given: 504

Intern
Joined: 22 Jun 2013
Posts: 40

Kudos [?]: 50 [0], given: 132

Re: When the positive integer x divided by the positive integer [#permalink]

### Show Tags

18 Jun 2014, 06:38
For me its E
i.e. All 3 values are possible

Given :
1. x = 3y + z

From here we can interpret that z < y .

2. Now given that when z is divided by y the remainder is 2.

Since z < y , i.e. Numerator is smaller that Denominator
Therefore we can safely assume the smallest possible value of z to be 2. (As we have to find the values that could be possible )

So x=3y + 2

Given Y is a +ve Integer
So checking for given Values which give us a Integer value of y

For 5 , y=1
For 8 , y= 2
For 32 , y=10

All satisfy the Equation.

Kudos [?]: 50 [0], given: 132

Math Expert
Joined: 02 Sep 2009
Posts: 43296

Kudos [?]: 139200 [0], given: 12779

Re: When the positive integer x divided by the positive integer [#permalink]

### Show Tags

22 Jun 2014, 03:14
Expert's post
2
This post was
BOOKMARKED
SOLUTION

When the positive integer is $$x$$ divided by the positive integer $$y$$, the quotient is 3 and the remainder is $$z$$. When $$z$$ is divided by $$y$$, the remainder is 2. Which of the following could be the value of $$x$$?

I. 5
II. 8
III. 32

A. I only
B. II only
C. III only
D. I and II only
E. I, II and III

When $$x$$ is divided by $$y$$, the quotient is 3 and the remainder is $$z$$: $$x=3y+z$$, where $$0\leq{z}<y$$ (the remainder must be less than the divisor).

When $$z$$ is divided by $$y$$, the remainder is 2: when divisor (y in our case) is more than dividend (z in our case), then the reminder equals to the dividend (for example, 2 divided by 5 gives the remainder of 2). Therefore, $$z=2$$ and $$2<y$$.

So, we have that $$x=3y+2$$ and $$2<y$$. This implies that the least value of $$x$$ is $$x=3*3+2=11$$: $$x$$ cannot be 5 or 8.

Could $$x$$ be 32? Yes. If $$y=10$$, then $$x=3*10+2=32$$.

Kudos points given to correct solutions above.

Try NEW remainders DS question.
_________________

Kudos [?]: 139200 [0], given: 12779

Manager
Joined: 21 Sep 2012
Posts: 218

Kudos [?]: 212 [0], given: 31

Location: United States
Concentration: Finance, Economics
Schools: CBS '17
GPA: 4
WE: General Management (Consumer Products)
Re: When the positive integer x divided by the positive integer [#permalink]

### Show Tags

22 Jun 2014, 04:39
niyantg wrote:
For me its E
i.e. All 3 values are possible

Given :
1. x = 3y + z

From here we can interpret that z < y .
2. Now given that when z is divided by y the remainder is 2.

Since z < y , i.e. Numerator is smaller that Denominator
Therefore we can safely assume the smallest possible value of z to be 2. (As we have to find the values that could be possible )

So x=3y + 2

Given Y is a +ve Integer
So checking for given Values which give us a Integer value of y

For 5 , y=1
For 8 , y= 2
For 32 , y=10

All satisfy the Equation.

As per above two highlighted statements, y can't take value of 1 or 2. y>2. Only 32 can satisfy the given conditions.

Kudos [?]: 212 [0], given: 31

Intern
Joined: 27 Apr 2013
Posts: 14

Kudos [?]: [0], given: 10

Concentration: General Management, Operations
GPA: 3.97
Re: When the positive integer x divided by the positive integer [#permalink]

### Show Tags

16 Jul 2014, 11:31
Bunnel, Z can be 1 also, right ?

Kudos [?]: [0], given: 10

Math Expert
Joined: 02 Sep 2009
Posts: 43296

Kudos [?]: 139200 [0], given: 12779

Re: When the positive integer x divided by the positive integer [#permalink]

### Show Tags

16 Jul 2014, 11:38
dheeraj787 wrote:
Bunnel, Z can be 1 also, right ?

No.

We are told that when z is divided by y, the remainder is 2 and we know also that y>z. Now, if z=1, then the reminder would be 1, not 2.

When z is divided by y, the remainder is 2: when divisor (y in our case) is more than dividend (z in our case), then the reminder equals to the dividend (for example, 2 divided by 5 gives the remainder of 2). Therefore, z=2.

Hope it's clear.
_________________

Kudos [?]: 139200 [0], given: 12779

Manager
Joined: 23 Sep 2015
Posts: 90

Kudos [?]: 20 [0], given: 213

Concentration: General Management, Finance
GMAT 1: 680 Q46 V38
GMAT 2: 690 Q47 V38
GPA: 3.5
Re: When the positive integer x divided by the positive integer [#permalink]

### Show Tags

05 Nov 2015, 17:53
Bunuel wrote:
SOLUTION

When the positive integer is $$x$$ divided by the positive integer $$y$$, the quotient is 3 and the remainder is $$z$$. When $$z$$ is divided by $$y$$, the remainder is 2. Which of the following could be the value of $$x$$?

I. 5
II. 8
III. 32

A. I only
B. II only
C. III only
D. I and II only
E. I, II and III

When $$x$$ is divided by $$y$$, the quotient is 3 and the remainder is $$z$$: $$x=3y+z$$, where $$0\leq{z}<y$$ (the remainder must be less than the divisor).

When $$z$$ is divided by $$y$$, the remainder is 2: when divisor (y in our case) is more than dividend (z in our case), then the reminder equals to the dividend (for example, 2 divided by 5 gives the remainder of 2). Therefore, $$z=2$$ and $$2<y$$.

So, we have that $$x=3y+2$$ and$$2<y$$. This implies that the least value of $$x$$ is $$x=3*3+2=11$$: $$x$$ cannot be 5 or 8.

Could $$x$$ be 32? Yes. If $$y=10$$, then $$x=3*10+2=32$$.

Try NEW remainders DS question.

Why is 2<y for x = 3y+2 and not 2<3y ?

This is the only part i cannot grasp.

Kudos [?]: 20 [0], given: 213

Math Expert
Joined: 02 Sep 2009
Posts: 43296

Kudos [?]: 139200 [0], given: 12779

Re: When the positive integer x divided by the positive integer [#permalink]

### Show Tags

05 Nov 2015, 19:56
GMATDemiGod wrote:
Bunuel wrote:
SOLUTION

When the positive integer is $$x$$ divided by the positive integer $$y$$, the quotient is 3 and the remainder is $$z$$. When $$z$$ is divided by $$y$$, the remainder is 2. Which of the following could be the value of $$x$$?

I. 5
II. 8
III. 32

A. I only
B. II only
C. III only
D. I and II only
E. I, II and III

When $$x$$ is divided by $$y$$, the quotient is 3 and the remainder is $$z$$: $$x=3y+z$$, where $$0\leq{z}<y$$ (the remainder must be less than the divisor).

When $$z$$ is divided by $$y$$, the remainder is 2: when divisor (y in our case) is more than dividend (z in our case), then the reminder equals to the dividend (for example, 2 divided by 5 gives the remainder of 2). Therefore, $$z=2$$ and $$2<y$$.

So, we have that $$x=3y+2$$ and$$2<y$$. This implies that the least value of $$x$$ is $$x=3*3+2=11$$: $$x$$ cannot be 5 or 8.

Could $$x$$ be 32? Yes. If $$y=10$$, then $$x=3*10+2=32$$.

Try NEW remainders DS question.

Why is 2<y for x = 3y+2 and not 2<3y ?

This is the only part i cannot grasp.

Check highlighted part.
_________________

Kudos [?]: 139200 [0], given: 12779

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7862

Kudos [?]: 18450 [1], given: 237

Location: Pune, India
Re: When the positive integer x divided by the positive integer [#permalink]

### Show Tags

05 Nov 2015, 21:19
1
KUDOS
Expert's post
Bunuel wrote:

When the positive integer $$x$$ is divided by the positive integer $$y$$, the quotient is 3 and the remainder is $$z$$. When $$z$$ is divided by $$y$$, the remainder is 2. Which of the following could be the value of $$x$$?

I. 5
II. 8
III. 32

A. I only
B. II only
C. III only
D. I and II only
E. I, II and III

Kudos for a correct solution.

This question is a great example of how understanding the concept of grouping for division can help you solve tricky questions easily.

x divided by y gives quotient 3 and remainder z.
What I see is x split into 3 groups having y balls each and z balls lying separately (these form the remainder)

When z is divided by y, remainder is 2.
z was the remainder when x was divided by y. So z is less than y. So when z is divided by y, the quotient will be 0 and the remainder will be z only. This means that z is 2. So now my figure morphs into x divided into 3 groups of y balls each with 2 balls lying on the side. The 3 groups will have more than 2 balls each, of course.

x = 3y + 2 where y is greater than 2.
So x cannot be 5 or 8. It can be 32 because y = 10 in that case.

To understand the concept of grouping, check:
http://www.veritasprep.com/blog/2011/04 ... unraveled/
http://www.veritasprep.com/blog/2011/04 ... y-applied/
http://www.veritasprep.com/blog/2011/05 ... emainders/
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for \$199

Veritas Prep Reviews

Kudos [?]: 18450 [1], given: 237

Board of Directors
Status: QA & VA Forum Moderator
Joined: 11 Jun 2011
Posts: 3224

Kudos [?]: 1174 [0], given: 327

Location: India
GPA: 3.5
WE: Business Development (Commercial Banking)
Re: When the positive integer x divided by the positive integer [#permalink]

### Show Tags

06 Nov 2015, 11:09
Bunuel wrote:
When the positive integer $$x$$ is divided by the positive integer $$y$$, the quotient is 3 and the remainder is $$z$$

$$\frac{x}{y} = 3 + z$$

Bunuel wrote:
When $$z$$ is divided by $$y$$, the remainder is 2.

$$\frac{z}{y} = k + 2$$

Bunuel wrote:
Which of the following could be the value of $$x$$?

Now lets start plugging in the options -

I. 5 - No possible satisfying conditions can be found out.
II. 8 - No possible satisfying conditions can be found out.

III. 32 - Lets try

$$\frac{32}{y} = 3 + z$$
$$\frac{32}{9} = 3(Quotient) + 5(Remainder)$$

Lets try the second step now -

$$\frac{z}{y} = k + 2$$

$$\frac{5}{y} = k +2$$

$$\frac{5}{3} = 1 (Quotient) +2(Remainder)$$

Only III Satisfices th condition m hence aswer is (C)
_________________

Thanks and Regards

Abhishek....

PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS

How to use Search Function in GMAT Club | Rules for Posting in QA forum | Writing Mathematical Formulas |Rules for Posting in VA forum | Request Expert's Reply ( VA Forum Only )

Kudos [?]: 1174 [0], given: 327

Target Test Prep Representative
Status: Head GMAT Instructor
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 1820

Kudos [?]: 1044 [0], given: 5

Re: When the positive integer x divided by the positive integer [#permalink]

### Show Tags

09 Nov 2016, 10:26
Bunuel wrote:

When the positive integer $$x$$ is divided by the positive integer $$y$$, the quotient is 3 and the remainder is $$z$$. When $$z$$ is divided by $$y$$, the remainder is 2. Which of the following could be the value of $$x$$?

I. 5
II. 8
III. 32

A. I only
B. II only
C. III only
D. I and II only
E. I, II and III

We are given that when the positive integer x is divided by the positive integer y, the quotient is 3 and the remainder is z. When z is divided by y, the remainder is 2. To determine which of the Roman numeral choices can be x, we can test each choice.

I. x = 5

If x = 5, no positive integer y exists that can produce a quotient of 3 when 5 is divided by y.

For example:

5/1 produces a quotient of 5.

5/2 produces a quotient of 2.

5/3 or 5/4 produces a quotient of 1.

Thus x cannot be 5.

II. x = 8

If x = 8, no positive integer y exists that can produce a quotient of 3 when 8 is divided by y.

For example:

8/1 produces a quotient of 8.

8/2 produces a quotient of 4.

8/3 or 8/4 produces a quotient of 2.

8/5, 8/6, 8/7, or 8/8 produces a quotient of 1.

Thus x cannot be 8.

III. x = 32

We see that x can be 32 since 32/10 produces a quotient of 3 and a remainder of 2, and when 2 is divided by 10, the quotient is 0 and the remainder is 2.

_________________

Jeffery Miller
Head of GMAT Instruction

GMAT Quant Self-Study Course
500+ lessons 3000+ practice problems 800+ HD solutions

Kudos [?]: 1044 [0], given: 5

Non-Human User
Joined: 09 Sep 2013
Posts: 14273

Kudos [?]: 291 [0], given: 0

Re: When the positive integer x divided by the positive integer [#permalink]

### Show Tags

23 Dec 2017, 11:27
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Kudos [?]: 291 [0], given: 0

Re: When the positive integer x divided by the positive integer   [#permalink] 23 Dec 2017, 11:27
Display posts from previous: Sort by

# When the positive integer x divided by the positive integer

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.