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The greatest common factor of 16 and the positive integer n

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Manager
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The greatest common factor of 16 and the positive integer n [#permalink]

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New post 09 Aug 2008, 18:42
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

The greatest common factor of 16 and the positive integer n is 4, and the greatest common factor of n and 45 is 3. Which of the following could be the greatest common factor of n and 210?

A 3
B 14
C 30
D 42
E 70

Experts, please help how I should approach this...thx
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Re: MGMAT GCF Problem [#permalink]

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New post 09 Aug 2008, 20:32
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"16 and the positive integer n is 4, and the greatest common factor of n and 45 is 3" implies that n has 3, and 4 as its factor. There may be some other factors but for sure 3 and 4 is there.

Moreover 45 = 3*3*5
So there is no more than one 3 and no 5 in n.

Similarly 16 = 4*4
So there is no more than one 4 in n.

We can say n = 12k

Question is asking what is GCF of 12k and 210.

210 = 2*3*5*7

We know 2 and 3 is already there in n, and also that n cannot have 5. But there is no restriction on 7. So n can have 7 as its factor.

So GCF = 2*3*7 = 42

Answer D.
Manager
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Re: MGMAT GCF Problem [#permalink]

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New post 09 Aug 2008, 21:23
Excellent! These type of problems always confuses me
+1 to you
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Re: MGMAT GCF Problem [#permalink]

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New post 10 Aug 2008, 06:01
kman wrote:
The greatest common factor of 16 and the positive integer n is 4, and the greatest common factor of n and 45 is 3. Which of the following could be the greatest common factor of n and 210?

A 3
B 14
C 30
D 42
E 70

Experts, please help how I should approach this...thx


good question :
My approach :
GCD between 16,n is 4 => n is a multiple of 5 => n=4*k k cannot be even number needs to be odd and can be 5,7,9,etc

GCD of n and 45 is 3 => n=3*m => m cannot be multiple of 3 or 5
m can be any other odd numbers like 7,19 etc and any even number such as 2,4,6,etc

now combine both conditions => n=4*3*L where L can be odd numbers except multiple of 3,5

hence lets factorise 210
210=5*3*2*7 already n has 2,3 and we know 5 cannot occur in factors of n hence largest possible GCD can be 3*2*7 =42
IMO D
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Re: MGMAT GCF Problem   [#permalink] 10 Aug 2008, 06:01
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