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Manager
Joined: 13 Dec 2005
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If the positive integer x is a multiple of 4?
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02 Dec 2006, 07:01
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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
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Re: If the positive integer x is a multiple of 4?
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05 May 2012, 03:40
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?
Thanks in advance! 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. Answer: B. Hope it's clear.
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Joined: 28 Dec 2005
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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



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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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Re: xy multiple.
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11 Sep 2011, 11:24
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.
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11 Sep 2011, 11:52
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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Re: xy multiple.
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11 Sep 2011, 15:08
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.
Answer is B.



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Re: xy multiple.
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12 Sep 2011, 02:36
simple way to answer these questions is to pick multiples of 4 and 6 and compare with answers.
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Re: xy multiple.
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15 Sep 2011, 13:49
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?
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05 May 2012, 02:11
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?
Thanks in advance!



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Re: If the positive integer x is a multiple of 4?
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07 May 2012, 16:52
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 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?
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11 Jul 2017, 18:56
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 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 7x 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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Re: If the positive integer x is a multiple of 4?
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30 Oct 2018, 00:23
Hi guys, i prepared a video with an explanation of this question, hope it is clear. Click here to see the video
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If the positive integer x is a multiple of 4?
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04 Oct 2019, 14:36
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?
Thanks in advance! 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. Answer: B. Hope it's clear. Hi BunuelI 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 !!



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Re: If the positive integer x is a multiple of 4?
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04 Oct 2019, 23:22
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?
Thanks in advance! 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. Answer: B. Hope it's clear. Hi BunuelI 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?
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Re: If the positive integer x is a multiple of 4?
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