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OG If x is a multiple of 4

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by AbeNeedsAnswers » Tue Jul 25, 2017 8:22 pm
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

B
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by Jay@ManhattanReview » Tue Jul 25, 2017 8:36 pm
AbeNeedsAnswers 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

B
Since x is a multiple of 4, we can write it as x = 4m; where m is an integer.

Similarly, since y is a multiple of 6, we can write it as y = 6n; where n is an integer.

=> xy = 4m*6n = 24mn

Let's take each of the three statements.

I. 8: Since xy = 24mn is completely divisible by 8, 'xy' is a multiple of '8.'
II. 12: Since xy = 24mn is completely divisible by 12, 'xy' is a multiple of '12.'
III 18: Since xy = 24mn may or may not divisible by 18; 'xy' is not necessarily a multiple of '18.'

The correct answer: C

Hope this helps!

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by Brent@GMATPrepNow » Wed Jul 26, 2017 6:25 am
AbeNeedsAnswers 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

B
A lot of integer property questions can be solved using prime factorization.
For questions involving divisibility, divisors, factors and multiples, we can say:
If N is divisible by k, then k is "hiding" within the prime factorization of N

Consider these examples:
24 is divisible by 3 because 24 = (2)(2)(2)(3)
Likewise, 70 is divisible by 5 because 70 = (2)(5)(7)
And 112 is divisible by 8 because 112 = (2)(2)(2)(2)(7)
And 630 is divisible by 15 because 630 = (2)(3)(3)(5)(7)
-----------ONTO THE QUESTION-------------------------------
positive integer x is a multiple of 4
In other words, x is divisible by 4, which means 4 is hiding in the prime factorization of x
So, we can write: x = (2)(2)(?)(?)(?)(?).... [ASIDE: The ?'s represents other primes that COULD be in the prime factorization. However, the only part of the prime factorization that we are certain of is the two 2's]

positive integer y is a multiple of 6
In other words, y is divisible by 6, which means 6 is hiding in the prime factorization of y
So, we can write: y = (2)(3)(?)(?)(?)(?)....

This means xy = (2)(2)(2)(3)(?)(?)(?)(?)....

This tells us that xy is divisible by all products formed by any combination of (2)(2)(2)(3)
So, xy must be divisible by 2, 3, 4, 6, 8, 12, 24

Answer: B

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Brent
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by [email protected] » Wed Jul 26, 2017 6:48 pm
Hi AbeNeedsAnswers,

This question can be solved by TESTing VALUES.

We're told that X and Y are both POSITIVE INTEGERS, that X is a multiple of 4 and that Y is a multiple of 6. We're asked what (X)(Y) MUST be a multiple of (meaning every time, no matter what values we use for X and Y).

With Roman Numeral questions, it's often useful to use the Roman Numerals 'against' the prompt, but for this question I'm going to do a bit of 'brute force' work first to define the patterns involved.

X could be 4, 8, 12, 16, 20, etc.
Y could be 6, 12, 18, 24, 30, etc.

So (X)(Y) could be 24, 48, 72, 96, etc. Starting with the smallest possibility, notice that ALL the possible values are multiples of 24. That means that (X)(Y) would have to be a multiple of both 8 and 12 (since both of those numbers divide into any multiple of 24). 18 does NOT divide evenly into 24 though, so that option is not part of the correct answer.

Final Answer: B

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by Scott@TargetTestPrep » Mon Aug 14, 2017 1:11 pm
AbeNeedsAnswers 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

B
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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by Matt@VeritasPrep » Fri Aug 18, 2017 2:18 pm
If x is a multiple of m and y is a multiple of n, then x*y must be a multiple of the LCM of m and n and ALL FACTORS of that LCM. Here the LCM of 4 and 6 is 12, so that's our answer: all factors of 12.
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