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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

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.

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.

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. Answer (D)
_________________

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!

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

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21 Jun 2015, 03:02

bschooladmit wrote:

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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Share some Kudos, if my posts help you. Thank you !

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

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08 Jul 2015, 03:16

1

This post received KUDOS

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.

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.

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) _________________

Thanks and Regards

Abhishek....

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

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 Time taken: 1:20
_________________

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

Show Tags

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. Answer (D)

"Since b is a multiple of 8, b = 23y." I am assuming that "23y" is meant to read (2^3)y?

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.

Answer: B
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