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Greatest common factor

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Junior | Next Rank: 30 Posts
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by C » Thu Mar 11, 2010 10:23 pm
Is the integer z devisible by 6?

(1) The greatest common factor of z and 12 is 3.
(2) The greatest common factor of z and 15 is 15.


Can someone please sovle this one, and explain the basic rules and how to proceed?

Thanks!
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Source: — Data Sufficiency |
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by mircealitoiu » Thu Mar 11, 2010 11:05 pm
imo A
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by kstv » Thu Mar 11, 2010 11:14 pm
(2) The GCF of z & 15 is 15 (5x3), so z is a mulitple of 15. Maybe 30, 45 etc.
30 is divisible by 6 but not 45 . Not sufficient.

(1) The GCF of z & 12 (2 x 2x 3) is 3.
z does not have a factor 2 , in that case the GCF should be 6 not 3, cos 12 has a factor 2.
Take the example 45(3x3x5) and 12(2x2x3) the GCF is 3 not 6.
Sufficient to say z is not divisible by 6.

IMO A
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by dxgamez » Thu Mar 18, 2010 4:35 pm
Hi,

Can you explain why st 1 is sufficient? I'm confused. I was thinking this way:-

if Z is to be divisible by 6(3x2), it needs to have both 3 and 2 in the prime factorization.

1. gcf of z and 12 is 3

12 - 3 x 2 x 2
z - 3 x ? x ?...

can i say that z is a multiple of 3, like you said for st 2 (z is multiple of 15)?

so if z is a multiple of 3, then z may be 6 or z may be 9. which is insufficient for 1. This is what I'm thinking. Could you give another explanation to help me understand where I'm seeing things wrongly?
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by ssuarezo » Thu Mar 18, 2010 7:14 pm
Hi Kstv:

If I applied the same logic in (1) as in (2), I would have:

(2) The GCF of z & 15 is 15 (5x3), so z is a mulitple of 15. Maybe 30, 45 etc.
30 is divisible by 6 but not 45 . Not sufficient. -- This is correct, I agree

(1) The GCF of z & 12 (2 x 2x 3) is 3.
z does not have a factor 2 , in that case the GCF should be 6 not 3, cos 12 has a factor 2.
Take the example 45(3x3x5) and 12(2x2x3) the GCF is 3 not 6.
Sufficient to say z is not divisible by 6.

About (1), I could say that z is multiple of 3, which could be 12, or 15. 12 is divisible by 6 (has a factor of 2), but not 15, so not suficient.

What am I doing wrong??

Thanks,
Silvia
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by schumi_gmat » Thu Mar 18, 2010 7:15 pm
IF GCF of z and 12 is 3 means that z is not a multiple of 2 and hence cannot be divisible by 6.

IMO A
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by kstv » Thu Mar 18, 2010 8:51 pm
ssuarezo wrote:Hi Kstv:

If I applied the same logic in (1) as in (2), I would have:

About (1), I could say that z is multiple of 3, which could be 12, or 15. 12 is divisible by 6 (has a factor of 2), but not 15, so not suficient.

What am I doing wrong??

Thanks,
Silvia
See the q (1) The greatest common factor of z and 12 is 3.
but you are considering z = 12 this is wrong
even then if z was 12 the factor of 2 you mention will make the GCF 3 X 2 X 2
The Q stem 1 says the GCF is 3 so z cannot be an even no.
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