The integers m and p are such that 2 < m < p and m is : DS Archive
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# The integers m and p are such that 2 < m < p and m is

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Director
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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, 18:03
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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 .
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Re: DS- Integers m and p [#permalink]

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18 Jun 2007, 19:38
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, 19:40
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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18 Jun 2007, 19:46
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.
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18 Jun 2007, 19:50
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.
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19 Jun 2007, 01:53
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
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19 Jun 2007, 02:59
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?
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19 Jun 2007, 11:08
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
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20 Jun 2007, 11:44
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
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20 Jun 2007, 13:08
'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
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20 Jun 2007, 14:17
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
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21 Jun 2007, 00:45
Himalayan wrote:
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

Thanks Himalayan for catching this. Yes 'D' shouldn't be right then.
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21 Jun 2007, 06:12
A for me.

1) SUFF - agree with bkk145's explanation.
2) INSUFF consider m=10 p=15
21 Jun 2007, 06:12
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# The integers m and p are such that 2 < m < p and m is

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