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romeo
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Post subject: GMAT prep Math - GCF-LCM  Posted: Wed Sep 26, 2007 6:10 pm |
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The integers m and p are such that 2<m<p and m is not factor of p. If r is remainder when p is divided by m then 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.
Thanks in advance for detailled explanation.
:oops:
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shaji
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Post subject: Re: GMAT prep Math - GCF-LCM Posted: Thu Sep 27, 2007 12:33 am |
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Statement 1 implies that m & p are even integers; therfore the remainder when p/m , the remainder r always is an integer greater than 2. SUFFICIENT.
Statement 2 implies that r=1, SUFFICIENT.
The correct answer is D
romeo wrote: The integers m and p are such that 2<m<p and m is not factor of p. If r is remainder when p is divided by m then 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.
Thanks in advance for detailled explanation.
:oops:
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S
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Post subject: Re: GMAT prep Math - GCF-LCM Posted: Thu Sep 27, 2007 2:40 am |
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Let m=5, p=6 such that 2<m<p & their LCM is 30, then p/m = 6/5 = 1 R 1 . So R is not greater than 1.
Answe should be A.
comment ?
.
shaji wrote: Statement 1 implies that m & p are even integers; therfore the remainder when p/m , the remainder r always is an integer greater than 2. SUFFICIENT. Statement 2 implies that r=1, SUFFICIENT. The correct answer is D romeo wrote: The integers m and p are such that 2<m<p and m is not factor of p. If r is remainder when p is divided by m then 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.
Thanks in advance for detailled explanation.
:oops:
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shaji
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Post subject: Re: GMAT prep Math - GCF-LCM Posted: Thu Sep 27, 2007 12:04 pm |
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A typo error in my earlier post.
Statement 1 implies that m & p are even integers; therfore the remainder when p/m , is always an integer >=2. SUFFICIENT. A definitive YES
Statement 2 implies that r=1, SUFFICIENT. A definitive NO
The correct answer is D
S wrote: Let m=5, p=6 such that 2<m<p & their LCM is 30, then p/m = 6/5 = 1 R 1 . So R is not greater than 1. Answe should be A. comment ? . shaji wrote: Statement 1 implies that m & p are even integers; therfore the remainder when p/m , the remainder r always is an integer greater than 2. SUFFICIENT. Statement 2 implies that r=1, SUFFICIENT. The correct answer is D romeo wrote: The integers m and p are such that 2<m<p and m is not factor of p. If r is remainder when p is divided by m then 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.
Thanks in advance for detailled explanation.
:oops:
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arjun666
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Post subject: Posted: Thu Sep 27, 2007 5:09 pm |
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I think Answer is A.
Only 1st statement is sufficient, since both are even integers other than 2 and m is not a factor of p.
2nd statement says LCM is 30. Consider 5 and 6, then r=1 (r=1). Consider 6 and 10, then r = 4 (>1). Consider 6 and 15, then r = 3 (>1). This is insufficient.
Therefore answer should be A
Regards
Arjun
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shaji
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Post subject: Posted: Fri Sep 28, 2007 11:58 am |
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arjun666@msn.com wrote: I think Answer is A.
Only 1st statement is sufficient, since both are even integers other than 2 and m is not a factor of p.
2nd statement says LCM is 30. Consider 5 and 6, then r=1 (r=1). Consider 6 and 10, then r = 4 (>1). Consider 6 and 15, then r = 3 (>1). This is insufficient.
Therefore answer should be A
Regards
Arjun
The correct answer indeed is A. Statement 2 is insufficient as elegantly explained by U.
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romeo
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Post subject: Posted: Wed Oct 03, 2007 11:59 pm |
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Thanks guys, great discussion.
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rajkoneru
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Post subject: Explain Posted: Fri Oct 05, 2007 12:40 am |
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Can you please explain how r =1 using statement 2.
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Guest83
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Post subject: Posted: Tue Oct 09, 2007 10:39 pm |
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Raj,
If you let M = 5, P = 6....you will see that P / M = 1 r1
Therefore, R = 1.
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bss
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Post subject: The integers m and p are such that 2<m<p Posted: Sun Feb 24, 2008 4:20 pm |
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I doubt the answer of A.
By statement1 we have that m and p both are even numbers. So for example we have m=6 and p=10. This also satisfies the condition that both have only 2 as the common factor. Now the point here is:
If I divide 10 by 6 -- I get a remainder 4. Which is >1 which is fine, it satisfies the answer as being A.
HOwever if I divide 10 by 6 and then remove the common 2 value---- reduce the equation to 5 / 3, then I get remainder as 2. Which is still ok and remainder is >1
Now another example let us take m = 4 and p = 6. Here also m / p will give remainder as 2 which is >1.
But if I reduce the equation 6 / 4 to 3 / 2. Then I get a remainder as 1. which is NOT > 1
So the answer A is doubtful.
Can the tutors pl explain.
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shaji
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Post subject: Re: The integers m and p are such that 2<m<p Posted: Thu Feb 28, 2008 1:13 am |
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Good question!!!. Do NOT reduce fractions, if U do so , the remainders too get reduced!!!
bss wrote: I doubt the answer of A.
By statement1 we have that m and p both are even numbers. So for example we have m=6 and p=10. This also satisfies the condition that both have only 2 as the common factor. Now the point here is:
If I divide 10 by 6 -- I get a remainder 4. Which is >1 which is fine, it satisfies the answer as being A.
HOwever if I divide 10 by 6 and then remove the common 2 value---- reduce the equation to 5 / 3, then I get remainder as 2. Which is still ok and remainder is >1
Now another example let us take m = 4 and p = 6. Here also m / p will give remainder as 2 which is >1.
But if I reduce the equation 6 / 4 to 3 / 2. Then I get a remainder as 1. which is NOT > 1
So the answer A is doubtful.
Can the tutors pl explain.
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StaceyKoprince
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Post subject: Posted: Mon Mar 03, 2008 10:10 pm |
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Shaji is right in his last post - you can't reduce the fractions without changing the remainders. Remember that these numbers don't represent a fraction; you are asked to perform long division. Write it that way, not as a fraction.
Also, Shaji, for your original mistake, you could have spotted that something was wrong because your two statements contradicted each other. (For one, you calculated that the remainder was greater than 2 and for the other you calculated that the remainder equaled 1. Those two statements are contradictory. If you ever calculate results that make the two statements contradict, you have done something wrong.)
_________________
Stacey Koprince
Instructor
Director of Online Community
ManhattanGMAT
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shaji
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Post subject: Posted: Thu Mar 06, 2008 12:01 pm |
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skoprince wrote: Shaji is right in his last post - you can't reduce the fractions without changing the remainders. Remember that these numbers don't represent a fraction; you are asked to perform long division. Write it that way, not as a fraction.
Also, Shaji, for your original mistake, you could have spotted that something was wrong because your two statements contradicted each other. (For one, you calculated that the remainder was greater than 2 and for the other you calculated that the remainder equaled 1. Those two statements are contradictory. If you ever calculate results that make the two statements contradict, you have done something wrong.)
Thanks Indeed!!!.
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StaceyKoprince
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Post subject: Posted: Wed Mar 12, 2008 5:27 pm |
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| ManhattanGMAT Staff |
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Posts: 6069
Location: San Francisco
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You're welcome!
_________________
Stacey Koprince
Instructor
Director of Online Community
ManhattanGMAT
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