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Source: — Data Sufficiency |
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by shankar.ashwin » Fri Oct 07, 2011 4:07 am
Generally for any problem involving LCM,GCD I try to break down the numbers in the problem to prime factors.

Here, z/6 = Integer?

A) GCF of z and 12 is 3

Now, 12 = 2*2*3
and z = 3 *(something) (now we know z cannot contain any factor of 2 as if 2 was a factor, the GCF would have been 6 instead of 3, so z could be some number such as 3*7 or 3*11)

If z does not have 2 as a factor, there is no way it could be divisible by 6, Hence sufficient.

B) GCF of z and 15 is 15

Now, 15 = 3*5
and z = 3*5*(something)

Here the (something) could take any number, I can have that number as 2 in one case and can have 7 in another. if it were to have a 2, it would be div by 6 and if there are no 2 factor it wouldn't be div by 6. Hence we cant say.
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by VivianKerr » Fri Oct 07, 2011 6:53 am
Pick numbers. It makes the statements much easier.

This is a yes or no question.

1) If the GCF of z and 12 is 3, then what if z = 3, then the answer is no. z cannot be 6, because then 6 would be the GCF, z cannot be 12, because then 12 would be the GCF. SUFF.

2) If the GCF of z and 15 is 15, then what if z = 15, the answer is no. If z = 30, then the answer is yes. INSUFF.
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by sunilrawat » Fri Oct 07, 2011 7:35 am
seema19 wrote:Is the integer z divisible by 6?

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

Answer: A

Please explain how to solve pbms like this.
For z to be divisible by 6, it has to be divisible both by 2 and 3.

(1) GCD of 12= 3*2*2 and z is 3.
This means 2 is NOT a factor of z. So z is NOT divisible by 6.
SUFFICIENT

(2) GCD of 15= 3*5 and z is 15
This means apart from 3, z may or may not have 2 as a factor.
INSUFFICIENT
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by parul9 » Fri Oct 07, 2011 8:07 pm
seema19 wrote:Is the integer z divisible by 6?

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

Answer: A

Please explain how to solve pbms like this.
Statement 1: the HCF of Z and 12 is 3.
12 = 3 * 4
Z should be = 3* (odd number)
This answers that Z is not divisble by 6.
So, statement 1 is sufficient.

Statement 2: HCF of Z and 15 is 15.
so, z = 15 * (any number)
Based on this information, Z can be anything.
So, this statement is not sufficient.

Hence the answer is A.
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