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# If the positive integer x is a multiple of 4 and the

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Math Expert
Joined: 02 Sep 2009
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If the positive integer x is a multiple of 4 and the  [#permalink]

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16 Jul 2012, 03:42
3
23
00:00

Difficulty:

25% (medium)

Question Stats:

68% (01:04) correct 32% (01:04) wrong based on 1838 sessions

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If the positive integer x is a multiple of 4 and the positive integer y is a multiple of 6, then xy must be a multiple of which of the following?

I. 8
II. 12
III. 18

(A) II only
(B) I and II only
(C) I and III only
(D) II and III only
(E) I, II, and III

Diagnostic Test
Question: 23
Page: 23
Difficulty: 550

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Re: If the positive integer x is a multiple of 4 and the  [#permalink]

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20 Jul 2012, 02:47
7
5
SOLUTION

If the positive integer x is a multiple of 4 and the positive integer y is a multiple of 6, then xy must be a multiple of which of the following?

I. 8
II. 12
III. 18

(A) II only
(B) I and II only
(C) I and III only
(D) II and III only
(E) I, II, and III

First of all notice that we are asked: "$$xy$$ MUST be a multiple of which of the following", not COULD be a multiple.

$$x$$ is a multiple of 4 --> $$x=4m$$, for some positive multiple $$m$$, so $$x$$ could be: 4, 8, 12, ...
$$y$$ is a multiple of 6 --> $$y=6n$$, for some positive multiple $$n$$, so $$y$$ could be: 6, 12, 18, ...

So, $$xy=(4m)*(6n)=24mn$$, hence $$xy$$ is in any case a multiple of 24, which means it must be a multiple of 8 and 12, but not necessarily of 18. For example the least value of $$x$$ is 4 and the least value of $$y$$ is 6, so the least value of $$xy$$ is 24, which is a multiple of both 8 and 12, but not 18.

Hope it's clear.
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Re: If the positive integer x is a multiple of 4 and the  [#permalink]

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16 Jul 2012, 12:10
8
x=4a
y=6b
xy=24ab
where a,b are integers
Hence , xy will always be divisible by 8,12 .
##### General Discussion
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Re: If the positive integer x is a multiple of 4 and the  [#permalink]

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16 Jul 2012, 09:53
1
Could it be A?
X multiple of 4, least value = 4 similarly, least value of y=6

LCM of 4&6=12 hence, A
8 can't be unless, x=2(4) or y=2(6)
18 can't be uncless x=3(4) or y=3(6)
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Re: If the positive integer x is a multiple of 4 and the  [#permalink]

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16 Jul 2012, 13:07
1
Prime factors of X = 2,2
Prime factors of Y = 2,3

Prime facots of XY = 2,2,2,3..

8 = 2*2*2
12 = 3*2*2

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If the positive integer x is a multiple of 4 and the  [#permalink]

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27 Apr 2015, 04:55
1
3
Hi Guys,

Please find below a practice question based on similar lines as the OG question in this thread:

If $$a$$ is a multiple of 5, $$b$$ is a multiple of 8 and $$c$$ is a multiple of 10 with $$c$$ having no common factor other than 10 with the product $$ab$$, then $$\frac{a^3*b^2}{c^2}$$ must be a multiple of which of the following?
I. 20
II. 40
III. 160

(A) I only
(B) I & III only
(C) II & III only
(D) I & II only
(E) I, II & III

Please post your analysis along with the answers. The OA and the solution will be posted on May 1, 2015.

Regards
Harsh
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If the positive integer x is a multiple of 4 and the  [#permalink]

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Updated on: 01 Apr 2018, 21:26
2
1
EgmatQuantExpert wrote:
Hi Guys,

Please find below a practice question based on similar lines as the OG question in this thread:

If $$a$$ is a multiple of 5, $$b$$ is a multiple of 8 and $$c$$ is a multiple of 10 with $$c$$ having no common factor other than 10 with the product $$ab$$, then $$\frac{a^3*b^2}{c^2}$$ must be a multiple of which of the following?
I. 20
II. 40
III. 160

(A) I only
(B) I & III only
(C) II & III only
(D) I & II only
(E) I, II & III

Please post your analysis along with the answers. The OA and the solution will be posted on May 1, 2015.

Regards
Harsh

The correct answer is Option D

Given
We are given three numbers a, b, c such that a is a multiple of 5, b is a multiple of 8 and c is a multiple of 10 with c having no common factors with the product ab other than 10. We are asked to find $$\frac{(a^3*b^2)}{c^2}$$ is a multiple of which of 20, 40 and 160.

Approach
We are given that a is a multiple of 5, so a = 5x.
Since b is a multiple of 8, b = $$2^3$$y
Also, c is a multiple of 10. So, we can write c = 2*5z. Now, we are given that the only factor that c has in common with the product ab is 10 (which is 2*5). The important inference to draw from this information is that z contains no 2 or 5.

We will use the above expressions to find out if $$\frac{(a^3*b^2)}{c^2}$$ is a multiple of which of 20, 40 and 160.

Working Out
Substituting the values of a, b and c, we can write $$\frac{(a^3*b^2)}{c^2}$$ = $$\frac{((5x)^{3}*(2^{3}*y)^2)}{(2*5z)^2}$$ = $$\frac{(5*2^{4}*x^{3}*y^{2})}{z^2}$$ which can be simplified to $$\frac{(80x^{3}y^{2})}{z^2}$$ . So, we can say that $$\frac{(a^{3}*b^{2})}{c^2}$$ will definitely be a factor of 20 and 40.
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Originally posted by EgmatQuantExpert on 01 May 2015, 05:51.
Last edited by EgmatQuantExpert on 01 Apr 2018, 21:26, edited 1 time in total.
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Re: If the positive integer x is a multiple of 4 and the  [#permalink]

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01 May 2015, 05:58
5
2
Sometimes, students get confused between questions of the following 2 types:

Type 1 Sample Question:
If P and Q are positive integers, is the product PQ a multiple of 24?
(1) P is a multiple of 4
(2) Q is a multiple of 6

(Note: the question above is a PS question of Type 1)

Type 2 Sample Question:

Is the positive integer n a multiple of 24?

(1) n is a multiple of 4.
(2) n is a multiple of 6.

(OG QR2 DS 115. Open discussion of this question is available here: http://gmatclub.com/forum/is-the-positive-integer-n-a-multiple-of-109886.html)

When these two types of Questions are put side by side, it’s easy to see the difference between them:

In Type 1 questions, the information is given about two distinct integers and the question is asked about the product of these two integers.

In order to answer the Sample Type 1 question given here, here’s what we should do:

From St. 1, P is of the form 4k
And from St. 2, Q is of the form 6m
So, PQ is of the form 4k*6m = 24*something.
Therefore, by combining the two statements, we are able to answer the question.

In Type 2 questions, both pieces of information are about the same integer n, and the question is also asked about n.

In order to answer the Sample Type 2 question given here, here’s what we should do:

From St. 1, n is of the form 4k. Prime factorize this. We get: n = $$2^{2}k$$
From St. 2, n is of the form 6m. Prime factorize this. We get: n = 2*3m
Combining the two statements, we see that n is of the form 2^{2}*3*something.

So, we can say for sure that n is divisible by 12 but cannot be definitive about divisibility by 24.

Therefore, the correct answer is E.

The mistake that some students make is that they solve Type 2 questions by using the approach for Type 1 questions. Please beware of this mistake!

Hope this discussion was useful for you!

Best Regards

Japinder
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Re: If the positive integer x is a multiple of 4 and the  [#permalink]

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21 Jun 2015, 03:02
1
Could it be A?
X multiple of 4, least value = 4 similarly, least value of y=6

LCM of 4&6=12 hence, A
8 can't be unless, x=2(4) or y=2(6)
18 can't be uncless x=3(4) or y=3(6)

You can not count LCM this way if you have 2 variables, there sets don't overlap --> See MGMAT NUmber Properties Chapter 6 - Poblem Set Ex. 4&5
For your solution you need question like this --> X is a multiple of 4 and X is a multiple of 6 (One variable)
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If the positive integer x is a multiple of 4 and the  [#permalink]

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Updated on: 08 Jul 2015, 10:00
Hi Mo2men,

It looks like I completely misread that question (it goes to show that a tired brain can impact anyone's performance). Kudos for catching it.

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Originally posted by EMPOWERgmatRichC on 07 Jul 2015, 21:27.
Last edited by EMPOWERgmatRichC on 08 Jul 2015, 10:00, edited 1 time in total.
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Re: If the positive integer x is a multiple of 4 and the  [#permalink]

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08 Jul 2015, 03:16
1
EMPOWERgmatRichC wrote:
Hi All,

In these types of questions, it often helps to focus on the SMALLEST possible value that fits all of the given facts. You'll find that larger and larger multiples of this initial number MIGHT be a multiple of certain numbers, but the smallest number is NOT a multiple of those numbers...

We're told that X and Y are positive integers, that X is a multiple of 4 and Y is a multiple of 6. We're asked what (X)(Y) MUST be a multiple of...

The smallest multiple of 4 and 6 is 12 (not 24). From here, it doesn't take much basic math to see that ALL multiples of 12 are multiples of 4 and 6, but we're going to focus on the number 12 since it's the smallest number that 'fits'

Is 12 a multiple of 8? NO.
Is 12 a multiple of 12? YES.
Is 12 a multiple of 18? NO.

There's only one answer that fits...

GMAT assassins aren't born, they're made,
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Hi Rich,
This question is #23 from diagnostic test in OG#13. Solution is B in OG13 page 55.

The question asks which is multiple of XY; I mean which is the multiple of the product of two numbers:

x=4k, y=6m so xy=24km

8 & 12 is multiple of 24

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If the positive integer x is a multiple of 4 and the  [#permalink]

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03 Dec 2016, 22:16
Nice Official Question.
Here is what i did in this Question =>

x=4k
y=6k'
so xy=24k*k"=24z for some integer z.
So xy is divisible by 24.
RULE => A number is divisible by factors as well as factors of its factors.
Notice 8 and 12 are both factors of 24 and 18 is not.
Hence 1 and 2 only.

Hence B

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Re: If the positive integer x is a multiple of 4 and the  [#permalink]

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04 Dec 2016, 00:23
Bunuel wrote:
If the positive integer x is a multiple of 4 and the positive integer y is a multiple of 6, then xy must be a multiple of which of the following?

I. 8
II. 12
III. 18

(A) II only
(B) I and II only
(C) I and III only
(D) II and III only
(E) I, II, and III

Let x = 4 , y = 6 ; so xy = 24 ( Multiple of 8 & 12 )
Let x = 8 , y = 6 ; so xy = 48 ( Multiple of 8 & 12 )
Let x = 8 , y = 12 ; so xy = 96 ( Multiple of 8 & 12 )

Try other combinations if you are not convinced , all are Multiple of 8 & 12 , hence, correct answer will be (B)

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Re: If the positive integer x is a multiple of 4 and the  [#permalink]

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18 Apr 2017, 04:16
As per statement
x=4,8,12
y=6,12,18
Take two values ox xy
xy=24,96
both can be evenly divided by 8,12 only
So Option: B
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If the positive integer x is a multiple of 4 and the  [#permalink]

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20 Aug 2017, 05:38
EgmatQuantExpert wrote:
EgmatQuantExpert wrote:
Hi Guys,

Please find below a practice question based on similar lines as the OG question in this thread:

If $$a$$ is a multiple of 5, $$b$$ is a multiple of 8 and $$c$$ is a multiple of 10 with $$c$$ having no common factor other than 10 with the product $$ab$$, then $$\frac{a^3*b^2}{c^2}$$ must be a multiple of which of the following?
I. 20
II. 40
III. 160

(A) I only
(B) I & III only
(C) II & III only
(D) I & II only
(E) I, II & III

Please post your analysis along with the answers. The OA and the solution will be posted on May 1, 2015.

Regards
Harsh

The correct answer is Option D

Given
We are given three numbers a, b, c such that a is a multiple of 5, b is a multiple of 8 and c is a multiple of 10 with c having no common factors with the product ab other than 10. We are asked to find $$\frac{(a^3*b^2)}{c^2}$$ is a multiple of which of 20, 40 and 160.

Approach
We are given that a is a multiple of 5, so a = 5x.
Since b is a multiple of 8, b = 23y
Also, c is a multiple of 10. So, we can write c = 2*5z. Now, we are given that the only factor that c has in common with the product ab is 10 (which is 2*5). The important inference to draw from this information is that z contains no 2 or 5.

We will use the above expressions to find out if $$\frac{(a^3*b^2)}{c^2}$$ is a multiple of which of 20, 40 and 160.

Working Out
Substituting the values of a, b and c, we can write $$\frac{(a^3*b^2)}{c^2}$$ = $$\frac{((5x)^{3}*(2^{3}*y)^2)}{(2*5z)^2}$$ = $$\frac{(5*2^{4}*x^{3}*y^{2})}{z^2}$$ which can be simplified to $$\frac{(80x^{3}y^{2})}{z^2}$$ . So, we can say that $$\frac{(a^{3}*b^{2})}{c^2}$$ will definitely be a factor of 20 and 40.

"Since b is a multiple of 8, b = 23y." I am assuming that "23y" is meant to read (2^3)y?
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Re: If the positive integer x is a multiple of 4 and the  [#permalink]

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13 Sep 2017, 03:21
"Since b is a multiple of 8, b = 23y." I am assuming that "23y" is meant to read (2^3)y?[/quote]

Prime factors of X = 2,2
Prime factors of Y = 2,3

Prime facots of XY = 2,2,2,3.. ->24

8 = 2*2*2
12 = 3*2*2

Only 8 and 12 can satisfy this
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Re: If the positive integer x is a multiple of 4 and the  [#permalink]

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20 Dec 2017, 11:32
Bunuel wrote:
If the positive integer x is a multiple of 4 and the positive integer y is a multiple of 6, then xy must be a multiple of which of the following?

I. 8
II. 12
III. 18

(A) II only
(B) I and II only
(C) I and III only
(D) II and III only
(E) I, II, and III

Since x is a multiple of 4, x = 4k for some positive integer k. Similarly, since y is a multiple of 6, y = 6m for some positive integer m. Thus, xy = (4k)(6m) = 24km. We see that regardless what k and m are, 24km will always be divisible by 8 and 12, but not necessarily by 18.

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Re: If the positive integer x is a multiple of 4 and the  [#permalink]

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28 Mar 2018, 10:18
EgmatQuantExpert wrote:
EgmatQuantExpert wrote:
Hi Guys,

Please find below a practice question based on similar lines as the OG question in this thread:

If $$a$$ is a multiple of 5, $$b$$ is a multiple of 8 and $$c$$ is a multiple of 10 with $$c$$ having no common factor other than 10 with the product $$ab$$, then $$\frac{a^3*b^2}{c^2}$$ must be a multiple of which of the following?
I. 20
II. 40
III. 160

(A) I only
(B) I & III only
(C) II & III only
(D) I & II only
(E) I, II & III

Please post your analysis along with the answers. The OA and the solution will be posted on May 1, 2015.

Regards
Harsh

The correct answer is Option D

Given
We are given three numbers a, b, c such that a is a multiple of 5, b is a multiple of 8 and c is a multiple of 10 with c having no common factors with the product ab other than 10. We are asked to find $$\frac{(a^3*b^2)}{c^2}$$ is a multiple of which of 20, 40 and 160.

Approach
We are given that a is a multiple of 5, so a = 5x.
Since b is a multiple of 8, b = 23y
Also, c is a multiple of 10. So, we can write c = 2*5z. Now, we are given that the only factor that c has in common with the product ab is 10 (which is 2*5). The important inference to draw from this information is that z contains no 2 or 5.

We will use the above expressions to find out if $$\frac{(a^3*b^2)}{c^2}$$ is a multiple of which of 20, 40 and 160.

Working Out
Substituting the values of a, b and c, we can write $$\frac{(a^3*b^2)}{c^2}$$ = $$\frac{((5x)^{3}*(2^{3}*y)^2)}{(2*5z)^2}$$ = $$\frac{(5*2^{4}*x^{3}*y^{2})}{z^2}$$ which can be simplified to $$\frac{(80x^{3}y^{2})}{z^2}$$ . So, we can say that $$\frac{(a^{3}*b^{2})}{c^2}$$ will definitely be a factor of 20 and 40.

Hello there EgmatQuantExpert are these the formulas you are using ? or what?
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If the positive integer x is a multiple of 4 and the  [#permalink]

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01 Apr 2018, 20:52
1
dave13 wrote:
EgmatQuantExpert wrote:
EgmatQuantExpert wrote:
Hi Guys,

Please find below a practice question based on similar lines as the OG question in this thread:

If $$a$$ is a multiple of 5, $$b$$ is a multiple of 8 and $$c$$ is a multiple of 10 with $$c$$ having no common factor other than 10 with the product $$ab$$, then $$\frac{a^3*b^2}{c^2}$$ must be a multiple of which of the following?
I. 20
II. 40
III. 160

(A) I only
(B) I & III only
(C) II & III only
(D) I & II only
(E) I, II & III

Please post your analysis along with the answers. The OA and the solution will be posted on May 1, 2015.

Regards
Harsh

The correct answer is Option D

Given
We are given three numbers a, b, c such that a is a multiple of 5, b is a multiple of 8 and c is a multiple of 10 with c having no common factors with the product ab other than 10. We are asked to find $$\frac{(a^3*b^2)}{c^2}$$ is a multiple of which of 20, 40 and 160.

Approach
We are given that a is a multiple of 5, so a = 5x.
Since b is a multiple of 8, b = 23y
Also, c is a multiple of 10. So, we can write c = 2*5z. Now, we are given that the only factor that c has in common with the product ab is 10 (which is 2*5). The important inference to draw from this information is that z contains no 2 or 5.

We will use the above expressions to find out if $$\frac{(a^3*b^2)}{c^2}$$ is a multiple of which of 20, 40 and 160.

Working Out
Substituting the values of a, b and c, we can write $$\frac{(a^3*b^2)}{c^2}$$ = $$\frac{((5x)^{3}*(2^{3}*y)^2)}{(2*5z)^2}$$ = $$\frac{(5*2^{4}*x^{3}*y^{2})}{z^2}$$ which can be simplified to $$\frac{(80x^{3}y^{2})}{z^2}$$ . So, we can say that $$\frac{(a^{3}*b^{2})}{c^2}$$ will definitely be a factor of 20 and 40.

Hello there EgmatQuantExpert are these the formulas you are using ? or what?

Hey Dave,

Happy to help

These are not formulas.

Understand it like this:

If $$a$$ is a multiple of $$5$$, the possible values of a can be $$5, 10 , 15, 20$$.. and so on.

Which is $$5*1, 5*2, 5*3, 5*4$$ and so on.

Thus, we can write $$a$$ as $$5x$$, where $$x$$ is any positive integer.

For example: $$a = 40, x =8$$ ; if $$a = 100, x = 20$$ and so on.

This is a general representation of the number $$a$$ and not any formula.

In a similar way, we can write:

$$b$$ as $$8y$$, where $$y$$ is any positive integer, and

$$c$$ as $$10z$$, where $$z$$ is any positive integer.

We wrote a, b, and c in such a manner to reduce the complexity of the problem by denoting each of them in their general form.

Regards.
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Re: If the positive integer x is a multiple of 4 and the  [#permalink]

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01 Apr 2018, 21:36
Bunuel wrote:
If the positive integer x is a multiple of 4 and the positive integer y is a multiple of 6, then xy must be a multiple of which of the following?

I. 8
II. 12
III. 18

(A) II only
(B) I and II only
(C) I and III only
(D) II and III only
(E) I, II, and III

Diagnostic Test
Question: 23
Page: 23
Difficulty: 550

x = 2^2
y = 2*3

i.e. x*y = 2^2*2*3 = 2^2*3 = 24

i.e. x*y must be a Multiple of all factors of 24

8 and 12 are factors of 24 but 18 is not a factor of 24 hence

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Re: If the positive integer x is a multiple of 4 and the &nbs [#permalink] 01 Apr 2018, 21:36

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