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Common Factor

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by fightthegmat » Tue Sep 08, 2009 10:31 am
If k, m, and t are positive integers and k/6 + m/4 = t/12, do t and 12 have a common factor greater than 1 ?
(1) k is a multiple of 3.
(2) m is a multiple of 3.

Can you help with the explanation
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Re: Common Factor

by gmat579 » Tue Sep 08, 2009 12:25 pm
fightthegmat wrote:If k, m, and t are positive integers and k/6 + m/4 = t/12, do t and 12 have a common factor greater than 1 ?
(1) k is a multiple of 3.
(2) m is a multiple of 3.

Can you help with the explanation

IMO - A

simplifying the given equation we get 2k + 3m = t

The question asks us if 3 or 4 is a factor of 't'

From Stmt 1 - since k is a multiple of 3, 3 has to be a factor of 't' as 2k + 3m = t

From Stmt 2 - We know that m is mutiple of 3 but we dont know about k. So not sufficient.
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by tom4lax » Tue Sep 08, 2009 12:55 pm
I am not sure if I follow the logic 100%.

How do we get from the question stem that 3 or 4 needs to be a factor of 12?

I get the simplification to t = 2k + 3m

Statement I) t = 2(3k) + 3m
This to me means t = 6k + 3m

Statement II) t = 2k + 3(3m)
means t = 2k + 9m

Taking a quick glance and plugging in 1's for k & m, from statement 1 we get t = 9 and from 2 we get t = 11?? I dont think this is the correct way to look at it though, but it makes the most sense to me.

Whats the correct thought process here? (Sorry for detailed question, but these types of questions come up a lot and I want to be able to nail down the concept)
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Re: Common Factor

by fightthegmat » Tue Sep 08, 2009 7:00 pm
gmat579 wrote:
fightthegmat wrote:If k, m, and t are positive integers and k/6 + m/4 = t/12, do t and 12 have a common factor greater than 1 ?
(1) k is a multiple of 3.
(2) m is a multiple of 3.

Can you help with the explanation

IMO - A

simplifying the given equation we get 2k + 3m = t

The question asks us if 3 or 4 is a factor of 't'

From Stmt 1 - since k is a multiple of 3, 3 has to be a factor of 't' as 2k + 3m = t

From Stmt 2 - We know that m is mutiple of 3 but we dont know about k. So not sufficient.
Could you pls elaborate more on the logic, if statement 1 says k is a multiple of 3 and as to be a factor of k then why not statement which says m is a multiple of 3 so it can be a factor of t(accor to your logic)

The OA is A,but i am not clear with the explanation
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by bharathh » Tue Sep 08, 2009 9:09 pm
Take the original eqn and multiply by 12.

You get 2k + 3m = t

Now for t to share a factor with 12 we need to prove that it is the product of 2 and some number or 3 and some number (factors of 12)

Statement 1

If k is a multiple of 3, it can be expressed as k = 3n

so 2*3n + 3m = t

Factor 3 out

3(2n + m) = t

Sufficient as t can be expressed as the product of 3 and someone number. So it shares 3 as a common factor with 12.


Statement 2

We already know that 3m is divisible by 3. Stating that m is also divisible by 3 does not help us any. Regardless lets plug in m = 3n

2k + 3(3m) = t

Cannot do anything more. So we have no idea if t can be expressed as a product of 2 and a number or 3 and a number.

Since statement 1 is sufficient, A
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by tom4lax » Wed Sep 09, 2009 5:46 am
Brilliant, thanks so much. The step where we factor out the 3 to get to t = 3(2n+m) is what I was missing.

Very clear now.
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