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rohansherry
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DS int m & p 24 Aug 2009, 10:00
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0% (00:00) correct 100% (02:20) wrong based on 1 sessions

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pls explain



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DS int m & p 24 Aug 2009, 10:26
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hi my approach for B was:

if 30 is the LCM.

The no can be ( m& p) .................Discard 2,15 & 1, 30
(5, 6) & ( 3, 10 )


her if we divide p by M the remainder will be one always....so i the nas is no ...ans stmt suff........

Pls comment on this...
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gmatjon
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DS int m & p 24 Aug 2009, 10:34
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Very interesting problem, need explanation as well...
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madeinafrica
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DS int m & p 24 Aug 2009, 11:18
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very interesting question

Here is my approch:

statement 1: greatest common factor of m and p is 2: means that 2 is a factor of m and 2 is a factor of P; p and m are both even numbers.
m<p means that m cannot be p-1 (since p-1 is odd) so m<p-1
we also know that m is not a factor of p and we can thus conclude that the remainder will always be greater that 1 when we divide p by m

statement 2 :
we can take 2<5<6 : remainder 1 when 6 is divided by 5

2<10<15 remainder 5.

thus insufficient
the answer is A

I don't know if could have done this if I didn't see the answer.
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DS int m & p Updated on: 31 Aug 2009, 08:33
Originally posted by LenaA on 24 Aug 2009, 11:45.
Last edited by LenaA on 31 Aug 2009, 08:33, edited 1 time in total.
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Property:
two consecutive numbers do not have common factors other than 1.



Statement 1 is sufficient:
To prove that r>1, we should prove that r cannot be 0 or 1.
r is not zero since p is not a multiple of m.
so we have to prove that r cannot be 1.
If r=1 when p is divided by m then p=mk+1 (p is a multiple of m plus 1).
From statement 1, m and p have a common factor. m is a factor of mk and mk will be divisible by all factors of m. Butp = mk +1 is a consecutive number and cannot be divisible by any factors of mk. So if m and p have common factors, the remainder will never be 1.


Statement 2 is insufficient. LCM does not mean that the numbers have common factors.

Took me a lot of time...definetly not 2 minutes...so not sure if this is a good solution for GMAT.
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DS int m & p 24 Aug 2009, 13:02
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thanks guys i just didnt .....include 10, 15 as a combination in the LCM.....
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DS int m & p 24 Aug 2009, 23:41
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LenaA
Property:
two consecutive numbers do not have common factors other than 1.



Statement 1 is sufficient:
To prove that r>1, we should prove that r cannot be 0 or 1.
r is not zero since p is not a multiple of m.
so we have to prove that r cannot be 1.
If r=1 when p is divided by m then p=mk+1 (p is a multiple of m plus 1).
From statement 1, m and p have a common factor. mk is a factor of m and will be divisible by all factors of m. But p = mk +1 is a consecutive number and cannot be divisible by any factors of mk. So if m and p have common factors, the remainder will never be 1.


Statement 2 is insufficient. LCM does not mean that the numbers have common factors.

Took me a lot of time...definetly not 2 minutes...so not sure if this is a good solution for GMAT.
Show more

Well, I tried plugging in numbers. It worked with some of them while it didnt with others.
Heres how:
Keeping in mind the question and stmt 1,
it worked when m = 12, p = 20, remainder results in 2.
it didn't work when m = 14, p = 100, remainder results in 1.

Could someone explain me where am I going wrong here. :?
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bhanushalinikhil
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DS int m & p 25 Aug 2009, 22:01
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rohansherry
bhanushalinikhil
LenaA
Property:
two consecutive numbers do not have common factors other than 1.



Statement 1 is sufficient:
To prove that r>1, we should prove that r cannot be 0 or 1.
r is not zero since p is not a multiple of m.
so we have to prove that r cannot be 1.
If r=1 when p is divided by m then p=mk+1 (p is a multiple of m plus 1).
From statement 1, m and p have a common factor. mk is a factor of m and will be divisible by all factors of m. But p = mk +1 is a consecutive number and cannot be divisible by any factors of mk. So if m and p have common factors, the remainder will never be 1.


Statement 2 is insufficient. LCM does not mean that the numbers have common factors.

Took me a lot of time...definetly not 2 minutes...so not sure if this is a good solution for GMAT.

Well, I tried plugging in numbers. It worked with some of them while it didnt with others.
Heres how:
Keeping in mind the question and stmt 1,
it worked when m = 12, p = 20, remainder results in 2.
it didn't work when m = 14, p = 100, remainder results in 1.

HOW COME YOU ARE GETTING REMAINDER AS 1 ..its not possible here. Remainder is 2 ...and it would be greater than 1 every time

Could someone explain me where am I going wrong here. :?
Show more

I see my mistake here. Thanks. :)
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patrykgb
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DS int m & p 31 Aug 2009, 07:25
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just quick question... for ii)

there are following combinations for m and p
30 and 1 --> remainder 0
15 and 2 --> remainder 1
10 and 3 --> remainder 1
6 and 5 --> remainder 1

right? in all these cases the reminder is not r>1... so it is a sufficient condition for me to make that judgement. No?
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LenaA
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DS int m & p 31 Aug 2009, 08:42
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patrykgb
just quick question... for ii)

there are following combinations for m and p
30 and 1 --> remainder 0
15 and 2 --> remainder 1
10 and 3 --> remainder 1
6 and 5 --> remainder 1

right? in all these cases the reminder is not r>1... so it is a sufficient condition for me to make that judgement. No?
Show more


But you are considering only a selection of possible pairs of numbers. What about (15,10) ...the remainder will be greater than 1...?
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patrykgb
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DS int m & p 01 Sep 2009, 00:02
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right... i just started revising math... thanks for teaching me what an multiple is. lol



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