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easy ps - factors/multiple problem

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easy ps - factors/multiple problem

by dhlee922 » Tue Feb 28, 2012 7:55 pm
what does oa mean? the first post says, "check the oa". does it mean original answer?

in the gmat og 12, diagnostic

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

i sorta understand the solution, but taking 4 and 6 and it's prime factors, why cant the product xy be 12? my reasoning is that the prime factors of 4 are 2 and 2, and the prime factors of 6 are 2 and 3.

i would've thought the unique primes are 2 * 2 * 3, since one 2 is shared between 4 and 6. therefore the answer i arrived at was A

or i guess my question is, when do i use all the primes for a product and when do i remember not to double count?


thanks
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by krusta80 » Tue Feb 28, 2012 8:10 pm
dhlee922 wrote:what does oa mean? the first post says, "check the oa". does it mean original answer?

in the gmat og 12, diagnostic

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

i sorta understand the solution, but taking 4 and 6 and it's prime factors, why cant the product xy be 12? my reasoning is that the prime factors of 4 are 2 and 2, and the prime factors of 6 are 2 and 3.

i would've thought the unique primes are 2 * 2 * 3, since one 2 is shared between 4 and 6. therefore the answer i arrived at was A

or i guess my question is, when do i use all the primes for a product and when do i remember not to double count?


thanks
CORRECTION!!

The following is to be used when determining the LEAST COMMON FACTOR for an integer that is divisible by 4 and by 6, HOWEVER that is not what this question is asking for!!

Step 1: Break up all numbers that are factors of the product into the product of prime factors.

4 = 2*2
6 = 2*3

Step 2: Pool together 1 instance of every listed prime number that is shared by all factors

Shared factor -> 2

Step 3: Pool together all non-repeated prime factors

Non-shared -> 2,3

Step 4: Multiply together all factors from both groups

2*2*3 = 12

===============================

In this question, they are giving us the product xy, which MUST be a multiple of 24. Therefore, any choice that is a factor of 24 will also be a factor of xy.

Answer is B
Last edited by krusta80 on Tue Feb 28, 2012 8:24 pm, edited 1 time in total.
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by dhlee922 » Tue Feb 28, 2012 8:14 pm
i'm sorry krusta80, but that answer is wrong. it's the same logic i applied. i didnt write out the answer b/c i didnt want to confuse people and wanted to get ideas on why my method was incorrect
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by krusta80 » Tue Feb 28, 2012 8:25 pm
dhlee922 wrote:i'm sorry krusta80, but that answer is wrong. it's the same logic i applied. i didnt write out the answer b/c i didnt want to confuse people and wanted to get ideas on why my method was incorrect
Sorry for my sloppy initial response...I'm a bit rusty. :)

I've edited my response now to adjust to the specifics of the problem.
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by Brent@GMATPrepNow » Fri Jan 10, 2020 5:27 am
dhlee922 wrote:what does oa mean? the first post says, "check the oa". does it mean original answer?

in the gmat og 12, diagnostic

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
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

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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