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

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

Answer B

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.
Answer (D)

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: https://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

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)

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.

GMAT assassins aren't born, they're made,
Rich

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

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

Since x is a multiple of 4 and y is a multiple of 6, Let's say x= 4*a and y = 6*a where a,b are positive integers.

The product x*y = 4*a * 6*b = 24* a* b

We can say that the product xy is a multiple of 24 or multiple of any factor of 24.

Since 8 and 12 are the factors of 24 , xy must be a multiple of 8 and 12.

xy can be or cannot be a multiple of 18. We are not sure about that.

Answer: Option B

Thanks,
Clifin J Francis

Can someone briefly explain why you don't use the LCM of the two multiples?

 
­
The LCM of x and y isn't relevant to the question because we're specifically asked about xy, not about finding the least common multiple of x and y. We have that x is a multiple of 4 and y is a multiple of 6, so when you multiply these together you get that xy must be a multiple of 24.

­No sweat here. Look at primes:

­

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