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Number theory -2

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Number theory -2

by guerrero » Wed Apr 03, 2013 10:07 am
The LCM of three numbers is four times their GCF. Which of the following must be true of the numbers?

I. At least one of the numbers is odd.
II. Two of the three numbers must be same.
III. At least one number is the same as GCF.

A. I only
B. III only
C. I and III only
D. I and II only
E. II and III only


Kindly elaborate the approach .. regards!

OA B
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by GMATGuruNY » Wed Apr 03, 2013 10:48 am
guerrero wrote:The LCM of three numbers is four times their GCF. Which of the following must be true of the numbers?

I. At least one of the numbers is odd.
II. Two of the three numbers must be same.
III. At least one number is the same as GCF.

A. I only
B. III only
C. I and III only
D. I and II only
E. II and III only


Kindly elaborate the approach .. regards!

OA B
Try to plug in values that prove that I, II and III DON'T have to be true.

To disprove I -- which states that at least one number must be odd -- let the GCF = 2.
Since the LCM = 4(GCF), the LCM = 8.
In this case:
The greatest factor common to all three numbers must be 2.
The least value divisible by all 3 numbers must be 8.
Thus, the three numbers could be 2, 4 and 8.

Since none of the numbers here are odd, eliminate any answer choice that includes I (A, C, and D).
Since all 3 numbers are different, eliminate any remaining answer choice that includes II (E).

The correct answer is B.
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by vipulgoyal » Wed Apr 03, 2013 10:09 pm
thought other way, very well explained, but what if plug numbers 3,6, 12 then ans is C
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by GMATGuruNY » Wed Apr 03, 2013 10:47 pm
vipulgoyal wrote:thought other way, very well explained, but what if plug numbers 3,6, 12 then ans is C
Statement I: At least one of the numbers is odd.
This statement COULD be true, as you should have shown with 3, 6 and 12.
But the question stem asks which of three statements MUST be true.
Since 2, 4, and 8 satisfy the condition that LCM = 4(GCF), it does NOT have to be true that at least one of the numbers is odd.
Thus, we should eliminate any answer choice that includes statement I.
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by Anju@Gurome » Thu Apr 04, 2013 12:08 am
Plug in numbers is the best approach for this kind of problem.
As Mitch has already posted that, I am posting the general approach which is not necessary to solve this problem but it might help someone to clear up the concepts.

Let us assume that the LCM is L and the GCF is G.
Hence, L = 4G

Now, the numbers will be of the form: aG, bG, and cG (As they should be multiples of G), where a, b, and c are positive integers.

None of a, b, and c can be less than 1 as the GCF is G and none of them can be greater than 4 as the LCM is 4G. Also, a, b, and c must be factors of 4 as the LCM is 4G.

Hence, possible values of a, b, and c are 1, 2, and 4.
As the GCF is G and LCM is 4G, at least one of them must be G and at least one of them must be 4G. Otherwise GCF and LCM will change. Hence, III must be true.

Now, possible values of the three numbers are,
  • G, 2G, 4G ---> II need not be true.
    G, G, 4G
    G, 4G, 4G
If G is even, no number will be odd.
Hence, I need not be true.

The correct answer is B.
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