GMAT Prep question - really tough one - LCM, GCF factor etc.

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Sorry about the previous incomplete test.

The integers m and p are such that 2 is less than m and m is less than p. Also, 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.

The correct answer is A

Your help is appreciated.

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by ildude02 » Tue Jun 17, 2008 1:25 pm
This seems to be tricky question. Is the answerr A? Since GF is 2, both M and P should be multiples of 2 and taking some examples, I see, that the remiander is always > 1. At the same time I'm not 100% sure.

I hope if someone has a better explanation.

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by netigen » Tue Jun 17, 2008 1:30 pm
A tells us that the two numbers are even. So the remainder will always be a +ve even number.

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by ildude02 » Tue Jun 17, 2008 2:35 pm
netigen, any easy way to solve the second choice ?

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by netigen » Tue Jun 17, 2008 2:55 pm
From B:

2<m<p<30

First factor 30 as 2x3x5

Now figure out possible numbers less than 30 greater than 2 that have 30 as LCM (use the factors)

(1) 5 x 6 remainder is 1 so r=1
(2) 10 x 15 remainder is 5 so r=5 or r> 1

we have r=1 and r>1 so insufficient

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psriniva wrote:Sorry about the previous incomplete test.

The integers m and p are such that 2 is less than m and m is less than p. Also, 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.

The correct answer is A

Your help is appreciated.

Paddy
From statement (2) one thing can be inferred that both m and p are factors of 30.
The factors of 30 are : 1,2,3,5,6,10,15&30
Since m& p are both greater than 2, lets take the combinations possible from 3 onwards.

The possible pairs of numbers satisfying the conditions (m is not a factor of p and LCM of m&p is 30) are:
(3,10); (5,6); (6,10);(10,15).

Except for the pairs (3,10) & (5,6) all other pairs satisfy the requirements of the answer (Remainder r>1 when m divides p).

So the condition : "The least common multiple of m and p is 30." itself does not satisfy the requirements. Hence insuf.

So answer choice is (A).

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by maihuna » Sun May 17, 2009 1:38 am
Ian,
Can you please help us understand it better?
Regards
Charged up again to beat the beast :)

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by aj5105 » Sun May 17, 2009 2:00 am
m, p are integers; 2 < m < p ; p is not divisible by m. r is the remainder.

r > 1 : YES

r < 1 : NO

Statement 1 : Consider numbers such as 4-6, 6-8, 8-10. Their GCF is 2 and the remainder is 2.

Sufficient.

Statement 2: Consider 5-6. The remainder is 1. Consider 2-15. The remainder is >1.

InSufficient.

(A)

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by crackgmat007 » Sun May 31, 2009 8:58 pm
Statement 2: Consider 5-6. The remainder is 1. Consider 2-15. The remainder is >1.

Can you help me understand this pls?

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by Stuart@KaplanGMAT » Sun May 31, 2009 9:10 pm
crackgmat007 wrote:
Statement 2: Consider 5-6. The remainder is 1. Consider 2-15. The remainder is >1.

Can you help me understand this pls?
First, 2-15 don't work, since 2 < m < p

However, as stated in posts above, the pairs 5/6 and 10/15 both satisfy the condition in (2), i.e. the lowest common multiple of m and p is 30.

Now that we have "legal" numbers (i.e. they follow all the rules that we're given), we can plug them back into the original question, "does p/m have a remainder greater than 1?"

If m=5 and p=6, then p/m has a remainder of 1. Is 1 > 1? NO

If m=10 and p=15, then p/m has a remainder of 5. Is 5 > 1? YES

Since we can get both a YES and NO answer from statement (2), it's insufficient.
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by aj5105 » Sun May 31, 2009 9:16 pm
Consider m to be 5 and p to be 6. When we divide 6 by 5, the remainder is 1.

No consider m to be 6 and p to 10. When we divide 10 by 6, the remainder is 4. [Sorry, I made a mistake here]

Is it clear now why (B) is insufficient?


crackgmat007 wrote:
Statement 2: Consider 5-6. The remainder is 1. Consider 2-15. The remainder is >1.

Can you help me understand this pls?

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by ghacker » Wed Jun 03, 2009 8:15 pm
Statement A is sufficient , Why?

It is given that 2<m<p and from statement I we know that the GCD is 2
that tells us that we can write m as 2*K and p as 2*S where K and S are relatively prime but P>m hence S>K so S/K >1 that tells us r is >1

Statement B is insufficient , Why ?

We are told that THE LCM is 30 , then m and p can take all the factors of 30,with out violating the given conditions ,but then m=15 and p=30 then r=0
also if m=6 and p=30 then r>1 hence B is insufficient :lol:

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by lunarpower » Wed Jun 17, 2009 1:21 am
whew! this one is a serious cacophony of rephrasing and interpretation.

let's translate:
m is not a factor of p.
if m were a factor of p, then the remainder upon dividing p by m would be 0.

therefore, we can translate the above statement as follows:
"the remainder upon dividing p by m is not 0."
in other words, it's an integer greater than 0.

--

the question:
is r > 1 ?
here's a HIGH-LEVEL INTERPRETATION of this problem.

if the remainder WERE 1, then p would be 1 more than a multiple of m.
if this is the case, then p and m CANNOT have any common factors, other than 1. (this is so because all factors of m are factors of (p-1), which is a multiple of m; a number greater than 1 can't be a factor of both (p-1) and p, which are consecutive integers.)

therefore, if m and p have common factors, then the answer to this question is YES.

(note that the converse is not necessarily true: even if there are no common factors, the answer still could be yes. for instance, 17 divided by 6 leaves a remainder of 5, even though 17 and 6 have no common factors. but, if we can establish that there are common factors, then that's enough to show that the answer is Yes.)

--

statement (1)

if this is true, then m and p have the factor 2 in common, so, YES.
sufficient.

--

statement (2)

this doesn't tell you whether m and p have common factors.
if m = 5 and p = 6, for instance, then r = 1.
if m = 10 and p = 15, then r = 5, which is > 1.
insufficient.

ans (a)

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by lunarpower » Wed Jun 17, 2009 1:21 am
secondary solution: (because the primary solution above is fairly obnoxious)

JUST PLUG IN NUMBERS.


statement (1)

let's just PICK A WHOLE BUNCH OF NUMBERS WHOSE GCF IS 2 and watch what happens. let's try to make the numbers diverse.
say,
4 and 6
6 and 8
8 and 10
10 and 12
...
4 and 10
6 and 14
6 and 16
8 and 18
8 and 22
...
in all nine of these examples, the remainders are greater than 1. in fact, there is an obvious pattern, which is that they're all even, since the numbers in question must be even.

in fact, i just thought of this, which is a much nicer, more ground-level approach to statement one:
in statement 1, both m and p are even. therefore, the remainder is even, so it's greater than 1.

done.

sufficient.

--

statement (2)
just pick various numbers whose lcm is 30.
notice the numbers selected above:
5 and 6 --> remainder = 1
10 and 15 --> remainder = 5 > 1
insufficient.
Ron has been teaching various standardized tests for 20 years.

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by navalpike » Fri Jul 03, 2009 9:19 am
Or……..

Just notice that S1 says

“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.”

This simply means that both m & p are multiples of 2 (since their common factor is 2)
M=2k
P=2s

We actually know more than this such as m<p and m is not a factor of p, but that should not change our method.

P=m+r
2s=2k+r
2s-2k=r
2(s-k)=r

We know that r cannot be 0. This says that the remainder is a multiple of 2 and thus, r>1. Sufficient.

S2 Insufficient as shown above. (Although I would welcome a more algebraic approach rather than pick numbers for LCM 30. The first time I got the problem wrong because I failed to consider 10 & 6)