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The integers m and p are such that 2 < m < p and m is [#permalink]
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17 Jun 2007, 19:03
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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?
( i ) The greatest common factor of m and p is 2.
( ii ) the least common multiple of m and p is 30 .



VP
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Re: DS Integers m and p [#permalink]
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18 Jun 2007, 20:38
LM 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?
( i ) The greatest common factor of m and p is 2.
( ii ) the least common multiple of m and p is 30 .
I got A
(1) Greatest common factor of m & p is 2. We know that both p and m must be even. Given 2<m<p, the minimum difference between the two integers is 2. It is also given that m isn't a factor of p; thus, r must be greater than 1. SUFFICIENT.
(2) p = (30^x) * integer
m = (30^x) * integer
In this case, if you plug in the intergers, you can get anything. INSUFFICIENT.



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Re: DS Integers m and p [#permalink]
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18 Jun 2007, 20:40
LM wrote: The integers m and p are such that 2 < m <p> 1?
( i ) The greatest common factor of m and p is 2.
( ii ) the least common multiple of m and p is 30 .
Should be A.
m and p are both greater than 2 (according to the info).
consider (i)
m & p both are even numbers. the smallest of even numbers that satisfy (i) and the given information are 4 & 6. clearly remainder r > 1 for all cases.
consider (ii)
m=3, p=10
or
m=5, p=6
or
m=2, p=15
clearly, r is not greater than 1 in all cases.
please feel free to correct me.
Guys, it's not a poll, votes for A,B,C,D or E won't count. Please provide your answers with explanations so that we can have a healthy discussion.
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Intern
Joined: 03 Aug 2006
Posts: 22

I got D because
(i) agree with explantions
(ii) The nos. can only be:
3,10 or 5,6 and not 2,15 as 2<m<p
for both 3,10 and 5,6 remainder is 1 hence less than 2. sufficient
I could be completely wrong for the second statement.



VP
Joined: 10 Jun 2007
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ridhimagupta wrote: I got D because
(i) agree with explantions (ii) The nos. can only be: 3,10 or 5,6 and not 2,15 as 2<m<p for both 3,10 and 5,6 remainder is 1 hence less than 2. sufficient
I could be completely wrong for the second statement.
Least common multiple isn't what you described.
Say you have 30 and 40.
30 = 2 *3 *5
40 = 2^3 *5
Lcm = 2^3 * 3 * 5 = 8 * 15 = 120.



Director
Joined: 14 Jan 2007
Posts: 774

I would go with 'D'.
Stmt1: the p/m could be 6/4, 12/10, 22/6....each of them has a remainder > 1
So SUFF
Stmt2: LCM is 30, p/m could be 6/5, 10/3, 15/2, 30/10...each of them has remainder <= 1
So SUFF



Intern
Joined: 03 Aug 2006
Posts: 22

Stmt2: LCM is 30, p/m could be 6/5, 10/3, 15/2, 30/10...each of them has remainder <= 1
So SUFF
You meant 30/1 right????
This is what I am getting confused about  if the statement says 2<m<p then they have to positive integers greater than 2.
If the LCM is 30 then the only nos. which qualify with all these conditions have to be 5,6 and 3,10. In which case remainder is 1.
Am I missing something about LCMs here?



Senior Manager
Joined: 03 Jun 2007
Posts: 376

I feel it is D. Agree with statement 1. But for 2 we can only have number 5,6 and 10,3 and in either case the remainder is 1 which is not greater than 1. So D



Manager
Joined: 28 Aug 2006
Posts: 160

Agreed its D, Sine there are only 2 sets of numbers (3,10) and (5,6) that meet the coondition and their reminder is constant 1



Intern
Joined: 04 Feb 2007
Posts: 37

'C' IT IS....
Stmt1: m= 4, 6, 8, 10
p= 6, 8, 10, 12......
so remainder(p/m) could be 6/4=2, 8/4=0, 10/4=2....
INSUFF
Stmt2: m= 3, 5, 6, 10, 3, 5, 6, 15
p = 10, 6, 10, 30, 30, 30, 30.......(PAIR WISE).........LCM is 30
SO p/m could be 6/5=1, 10/3=1, 10/6=4, 30/10=0...
So INSUFF
(1)+(2): p/m=10/6, 10/30, 6/30 fetching r=0 all the time which is less than 1 all through.SUFF
C



Director
Joined: 26 Feb 2006
Posts: 899

D is not necessarily correct. what if m = 6 and p = 10?
http://www.gmatclub.com/phpbb/viewtopic ... =remainder
vshaunak@gmail.com wrote: I would go with 'D'.
Stmt1: the p/m could be 6/4, 12/10, 22/6....each of them has a remainder > 1 So SUFF
Stmt2: LCM is 30, p/m could be 6/5, 10/3, 15/2, 30/10...each of them has remainder <= 1 So SUFF



Director
Joined: 14 Jan 2007
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Himalayan wrote: D is not necessarily correct. what if m = 6 and p = 10? http://www.gmatclub.com/phpbb/viewtopic ... =remaindervshaunak@gmail.com wrote: I would go with 'D'.
Stmt1: the p/m could be 6/4, 12/10, 22/6....each of them has a remainder > 1 So SUFF
Stmt2: LCM is 30, p/m could be 6/5, 10/3, 15/2, 30/10...each of them has remainder <= 1 So SUFF
Thanks Himalayan for catching this. Yes 'D' shouldn't be right then.



Senior Manager
Joined: 11 Feb 2007
Posts: 351

A for me.
1) SUFF  agree with bkk145's explanation.
2) INSUFF consider m=10 p=15










