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GMATprep problem

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by sumatitandon » Fri Jan 09, 2009 3:05 am
The integers m and p are such that 2 < m< p , and m is not a factor of p, if r is the remainder , when p is divided by m , is r > 1?
1) The greatest common factor of m and p is 2
2) the least common factor of m and p is 30
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Source: — Data Sufficiency |

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Re: GMATprep problem

by Ian Stewart » Fri Jan 09, 2009 6:40 am
sumatitandon wrote:The integers m and p are such that 2 < m< p , and m is not a factor of p, if r is the remainder , when p is divided by m , is r > 1?
1) The greatest common factor of m and p is 2
2) the least common factor of m and p is 30
If 1) is true, then m and p are both divisible by 2; in other words, m and p are both even. Whenever you divide one even number by another, the remainder must be even, so r cannot be 1. Since we're told r is not zero, then r must be greater than 1, and this statement is sufficient.

If 2) is true, it is still possible that r = 1. For example, we might have p = 6 and m = 5.

A.
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confused

by shulapa » Fri Jan 09, 2009 9:22 am
Hi,

I am confused :oops: . I think I miss understood your analysis of the second statement. I think I don't understand the difference between common factor and common multiple.
I thought that 30 is the "common multiple" of 5 and 6. Actually I cannot see how can 30 can be the "least common factor".

Can you please clarify on the difference between the two terms.

Thanks,
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by cramya » Fri Jan 09, 2009 5:22 pm
Shulapa,
I think the 2nd statement should read least common multiple.

Regards,
Cramya
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by Ian Stewart » Fri Jan 09, 2009 5:31 pm
Yeah, I was remembering the problem from GMATPrep. The least common factor of two positive integers is always 1 - it could never be 30 - so the question doesn't make much sense as written in the original post. The original question reads 'least common multiple' in statement 2.
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by cramya » Fri Jan 09, 2009 5:41 pm
Thanks Ian!
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by sumatitandon » Fri Jan 09, 2009 6:21 pm
i m sorry.... stat 2 states lest common multiple.....

but why statement 2 is not sufficent ., i have tried other numbers also for ex: 3 and 10 the remainder is 1 again , so i feel that since remainder is 1 can't we conclude that r is not greater than 1 , pls explain where i m wrong......
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by cramya » Fri Jan 09, 2009 6:33 pm
m=5 p=6

The remainder is 1 not greater than 1

m=10 p=15

remainder >1

When the remainder is equal to 1 we get NO (i.e if m was 5 and p 6)

When the remainder > 1 we get a YES

Hence stmt II is INSUFF (we cant say for sure)


P.S: IMO when the question is " Is something greater than something" the equalt to case and lesser than case would fall in the NO bucket.

When a question reads " Is something lesser than something" then the equal to case and greater than case would fall in the NO bucket

Ian, please correct me if I am mistaken here.

Hope this helps!

Regards,
Cramya
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by Ian Stewart » Fri Jan 09, 2009 11:45 pm
cramya wrote:
P.S: IMO when the question is " Is something greater than something" the equalt to case and lesser than case would fall in the NO bucket.

When a question reads " Is something lesser than something" then the equal to case and greater than case would fall in the NO bucket

Ian, please correct me if I am mistaken here.
Yes, that's exactly right.
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by cramya » Fri Jan 09, 2009 11:58 pm
Thanks again Ian!!
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