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GCF practice

by logitech » Sat Jan 17, 2009 11:57 pm
Is z divisible by 6 ? (Z is an integer)

1) GCF(z;12)=3

1) GCF(z;15)=15
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Source: — Data Sufficiency |
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by DanaJ » Sun Jan 18, 2009 12:08 am
First one is sufficient. For z to be divisible by 6, it has to be divisible by both 2 and 3. (1) tells us that it is indeed divisible by 3, since 3 is one of its factors. But 2 is not, because, if 2 were one of its factors, that GCF(z, 12) would be at least 2*3 (because 12 is even).

A is the answer.
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by logitech » Sun Jan 18, 2009 12:11 am
Good job Danaj, I am sure you want to explain why second statement is not sufficient.
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by DanaJ » Sun Jan 18, 2009 12:14 am
Statement 2 only tells you that z is divisible by both 3 and 5. However, that does not shed light on whether z is divisible by 2.
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by mjjking » Tue Jan 20, 2009 1:38 pm
can you pls explain why b is different in reasoning from A??
After all, we have two GCF here... And 15 isn't divisible by 2.... so, where's my error?
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