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Great Questions! Need help, gmat tomo =(

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by houstonrockets16 » Sun Jun 28, 2009 5:44 pm
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 multiple of m and p is 30.

I am trying to find a way to answer this question besides just guessing numbers. Can anyone help? Thanks so much

OA: A
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Source: — Data Sufficiency |

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Re: Great Questions! Need help, gmat tomo =(

by Ian Stewart » Sun Jun 28, 2009 6:00 pm
houstonrockets16 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 multiple of m and p is 30.

I am trying to find a way to answer this question besides just guessing numbers. Can anyone help? Thanks so much

OA: A
From the stem, we know that p is not divisible by m, so the remainder can't be 0 when p is divided by m. So r is either equal to 1, or r is greater than 1.

According to Statement 1, m and p are both even (both are divisible by 2). When you divide an even number by another even number, the remainder can't be odd. Since the remainder is not zero, it must be 2 or greater. Sufficient.

Statement 2 is not sufficient, though you really need to look at examples to see why. We may have that m = 5, p = 6, or we may have that m = 10, p = 15.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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Re: Great Questions! Need help, gmat tomo =(

by sanjay_dce » Mon Jun 29, 2009 11:00 am
houstonrockets16 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 multiple of m and p is 30.

I am trying to find a way to answer this question besides just guessing numbers. Can anyone help? Thanks so much

OA: A
OA is correct infact this is a very trivial remainder theory concept..

when a number X is divided by number Y then the remainder left is always a multiple of HCF of X and Y ( provided x and y are integers) hence A is sufficient
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