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Manager  Joined: 13 Dec 2005
Posts: 59
If the positive integer x is a multiple of 4?  [#permalink]

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Question Stats: 67% (01:04) correct 33% (01:06) wrong based on 504 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

Official Guide 12 Question Question: 23 Page: 23 Difficulty: 600

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Video Explanations:
Math Expert V
Joined: 02 Sep 2009
Posts: 58320
Re: If the positive integer x is a multiple of 4?  [#permalink]

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4
6
andih wrote:
Hello,

quick question, I got the answer right, though I am having some problems with the "plug in numbers" approach here. Can anybody help real quick with an example?

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

So, $$xy=(4m)*(6n)=24mn$$, so $$xy$$ 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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Director  Joined: 28 Dec 2005
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1
2
nah, not necessarily:

Ans B.

From question x/4 is an integer.

x/4 = 2*2*q/4 ---> where q is the integer

Also y/6 = p => y/6 = 2*3*p/6

So, xy = 2*2*2*3 *pq

Thus 2*2*2 = 8 and 2*3*2 = 12 satisfy
Senior Manager  Joined: 20 Feb 2006
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B for me

x/4 = integer or 4p/4 = integer and y/6 = integer or yq/6 = integer

x and y are positive integers so limiting factor is when x and y are at theier minumum. Min values of x and y are when p=q=1
4*1 = 4 6*1 = 6
4*6 = 24

Hence multipe of 8 and 12
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GMAT 1: 780 Q51 V48 GRE 1: Q800 V740 Re: xy multiple.  [#permalink]

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xy = 4 * 6 = 2 * 2 * 2 * 3 = 8 * 3 = 2 * 12

We can see that xy is a multiple of both 8 and 12, but not 18.

Option (B)
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Re: xy multiple.  [#permalink]

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GyanOne wrote:
xy = 4 * 6 = 2 * 2 * 2 * 3 = 8 * 3 = 2 * 12

We can see that xy is a multiple of both 8 and 12, but not 18.

Option (B)

@GyanOne - LCM of 4 and 6 = 12 or if I compare as mentioned in the below series... I can see only multiple of 12.
Multiple of 4 - 4, 8, [highlight]12[/highlight], 16, 20, [highlight]24[/highlight]...
Multiple of 6 - 6, [highlight]12[/highlight], 18, [highlight]24[/highlight]...

Can you pls elaborate more?
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Director  Joined: 01 Feb 2011
Posts: 551
Re: xy multiple.  [#permalink]

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1
x is a multiple of 4 = 4p

y is a multiple of 6 = 6q

=> xy = 24pq

by quick observation we can see 8 and 12 are factors of 24. so i and iii must be true.

also 24 = 2*2*2*3

so xy may or may not have 18 as a factor depending on the values of p and q.

if p or q = 3 , then 18 is also a factor xy.
if p and q do not have 3, then 18 is not a factor or xy.

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Re: xy multiple.  [#permalink]

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simple way to answer these questions is to pick multiples of 4 and 6 and compare with answers.
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Manager  Joined: 08 Sep 2011
Posts: 53
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Re: xy multiple.  [#permalink]

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I did the prime factor box method:

4= 2*2
6= 2*3

so all factors of x has to have 2*2 in it and y has to have 2*3 in it.

lets look at our options

i) 8= 2*2*2. If we look at our prime factor boxes above we see that we have to have at least three 2's. (two in 4 and one in 6) so yes 8 is a multiple.

ii) 12= 2*2*3. Again if we look at our prime factor boxes above we see that it contains 2*2*3.

iii) 18= 2*3*3. For 18 we need two 3's and if you look at your prime factor boxes you see that we only have one 3, therefore 18 is not an option.

B both i and ii
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Re: If the positive integer x is a multiple of 4?  [#permalink]

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Hello,

quick question, I got the answer right, though I am having some problems with the "plug in numbers" approach here. Can anybody help real quick with an example?

Intern  Joined: 05 Apr 2012
Posts: 39
Re: If the positive integer x is a multiple of 4?  [#permalink]

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ellisje22 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

Official Guide 12 Question Question: 23 Page: 23 Difficulty: 600

Find All Official Guide Questions

Video Explanations:

Hello
I try !
X can be written as X=2x2K where K is a whole number which belongs to R
Y can be written as Y= 3X2 L where L is a whole number which belongs to R
HENCE XY can BE WRITTEN as 2x2x3x2xKxL

Hence XY must be a multiple of 8 and 12
For 18 we misse a 3

Hope I am right

best regards
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If the positive integer x is a multiple of 4?  [#permalink]

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ellisje22 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

Official Guide 12 Question Question: 23 Page: 23 Difficulty: 600

Find All Official Guide Questions

Video Explanations:

Theory,
All the below statements mean same:
7 is a factor of 21
7 is a divisor of 21
21 is a multiple of 7
21 is divisible by 7

x is multiple of 4
i.e. x=4p
y is a multiple of 6
i.e. y=6q
xy = 4p x 6q=24pq

The question stem asks: xy must be a multiple of which of the following?
i.e. Which of the following(8, 12, or 18) must be a divisor or a factor of 24pq?
i. 8
24/8=3 divisible
ii. 12
24/12=2, divisble
iii. 18
24/18=8/6=4/3, not divisble
Thus, Ans is B
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Manager  P
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Posts: 104
Re: If the positive integer x is a multiple of 4?  [#permalink]

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Hi guys, i prepared a video with an explanation of this question, hope it is clear.

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Manager  S
Joined: 15 Dec 2016
Posts: 100
If the positive integer x is a multiple of 4?  [#permalink]

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Bunuel wrote:
andih wrote:
Hello,

quick question, I got the answer right, though I am having some problems with the "plug in numbers" approach here. Can anybody help real quick with an example?

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

So, $$xy=(4m)*(6n)=24mn$$, so $$xy$$ 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.

Hi Bunuel

I selected (A)

Any idea on where my logic is wrong

-- i wrote down, x = 4 (given x is a multiple of 4) and y = 6 (given y is a multiple of 6)

-- For x = 4 and y =6 ; In this case, the Lowest common multiple between these two numbers is actually 12 { I set to calculate the L.C.M as i thought this will give me a multiple of those two numbers }

-- Thus i thought, a L.C.M of 12 for x =4 and L = 6 ....thus the multiple must be 12 | 24 | 36 | 48 | 60 | 72 ......

Hence i selected (A) as "8" is not a multiple of 12

Where is the break in my logic ?

Thank you !!
Math Expert V
Joined: 02 Sep 2009
Posts: 58320
Re: If the positive integer x is a multiple of 4?  [#permalink]

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jabhatta@umail.iu.edu wrote:
Bunuel wrote:
andih wrote:
Hello,

quick question, I got the answer right, though I am having some problems with the "plug in numbers" approach here. Can anybody help real quick with an example?

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

So, $$xy=(4m)*(6n)=24mn$$, so $$xy$$ 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.

Hi Bunuel

I selected (A)

Any idea on where my logic is wrong

-- i wrote down, x = 4 (given x is a multiple of 4) and y = 6 (given y is a multiple of 6)

-- For x = 4 and y =6 ; In this case, the Lowest common multiple between these two numbers is actually 12 { I set to calculate the L.C.M as i thought this will give me a multiple of those two numbers }

-- Thus i thought, a L.C.M of 12 for x =4 and L = 6 ....thus the multiple must be 12 | 24 | 36 | 48 | 60 | 72 ......

Hence i selected (A) as "8" is not a multiple of 12

Where is the break in my logic ?

Thank you !!

The question does is NOT ...then the least common multiple of x and y must be a multiple of which of the following?

The question is ...then xy (THE PRODUCT OF x AND y) must be a multiple of which of the following?
_________________ Re: If the positive integer x is a multiple of 4?   [#permalink] 04 Oct 2019, 23:22
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