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# The integers m and p are such that 2<m<p and m is not

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Manager
Joined: 13 Dec 2005
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The integers m and p are such that 2<m<p and m is not [#permalink]  06 Jan 2006, 08:05
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100% (02:43) correct 0% (00:00) wrong based on 1 sessions
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.

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
EACH statement ALONE is sufficient.
Statements (1) and (2) TOGETHER are NOT SUFFICIENT
Director
Joined: 24 Oct 2005
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Re: GMATPrep DS [#permalink]  06 Jan 2006, 13:00
ellisje22 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.

m<p.. and we know that m/p <> 0.. is r>1

in the first case.. the greatest common factor is 2.. so.. for example 4 and 6.. GCF = 2.. r> 1.. holds true for all the numbers..

second case.. the least multiple of m and p is 30..
m =5 p =6.. m 3 p=10.. r =1 in all cases..

so D
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Manager
Joined: 13 Dec 2005
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I guess I am confused by the question stem. Before getting either statement 1 or statement 2, you are told that both m and p are greater than 2. You also know that m is not a factor of p.

Given these two things...how is it EVER possible to have a remainder less than 1?
Manager
Joined: 15 Aug 2005
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IMO it should be A.

i. We know the two nos. are even, both nos. > 2 & one is not a factor of the other. This means that any multiple of the smaller number will be atleast 2 less than the greater no. - Sufficient

ii. consider two cases (m=5,p=6) and (m=6,p=10) both have LCM=30 but will have remainder 1 & 4 respectively - insuff
Manager
Joined: 15 Aug 2005
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ellisje22 wrote:
Given these two things...how is it EVER possible to have a remainder less than 1?

I think the question is really asking if you can be sure:
that the remainder is > 1 as opposed to remainder=1.
Manager
Joined: 13 Dec 2005
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The OA is A. This was from PowerPrep so unfortunately there is no OE. Thanks for your help!
Director
Joined: 24 Oct 2005
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believe2 wrote:
IMO it should be A.

i. We know the two nos. are even, both nos. > 2 & one is not a factor of the other. This means that any multiple of the smaller number will be atleast 2 less than the greater no. - Sufficient

ii. consider two cases (m=5,p=6) and (m=6,p=10) both have LCM=30 but will have remainder 1 & 4 respectively - insuff

yes u are rightt.. my idea of lcm was not right!!
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Director
Joined: 04 Oct 2005
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Damn i got it wrong..I thught it wud be D..
Apart from plugging in values, is there a technicue to solve it?
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