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The integers m and p are such that 2<m<p, and m is not a

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The integers m and p are such that 2<m<p, and m is not a [#permalink] New post 19 Sep 2010, 23:56
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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?

(1) the greatest common factor of m and p is 2
(2) the least common multiple of m and p is 30

--------------------------------------------------------------
A small doubt! Are the concepts GCD,LCM applicable to negative numbers?If so,please explain and tel me why -ve numbers are not considered in the Fact1 of the following problem?

--------------------------------------------------------------
I understood that Fact2 is not sufficient but am not sure about Fact1. If negative numbers are taken,then,for example, GCD(-8,6) is 2 and gives new remainder.Am I going wrong somewhere?
[Reveal] Spoiler: OA

Last edited by Bunuel on 06 Jun 2013, 03:29, edited 2 times in total.
Edited the question and added the OA.
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Re: DS:GCD of numbers. [#permalink] New post 20 Sep 2010, 01:20
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ravitejapandiri wrote:
A small doubt!Are the concepts GCD,LCM applicable to negative numbers?If so,please explain and tel me why -ve numbers are not considered in the Fact1 of the following problem?

--------------------------------------------------------------
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?

(1) the greatest common factor of m and p is 2
(2) the least common multiple of m and p is 30
--------------------------------------------------------------

I understood that Fact2 is not sufficient but am not sure about Fact1.If negative numbers are taken,then,for example, GCD(-8,6) is 2 and gives new remainder.Am I going wrong somewhere?


(1) This is sufficient. We know gcd(m,p)=2, this means both m & p are even.
Let q be the quotient and r the remainder when p is divided by m
p = q*m + r
Note that r cannot be 0, since if r=0, then m divides p and is therefore a factor of p, which we know is not true
Also note r cannot be 1, since if r=1, them q*m (even number) + 1 (odd) is an odd number, which is not possible since we know that p is even
Hence r>1.

(2) Not sufficient. eg. m=6,p=15,lcm=30, remainder=3; m=5,p=6,lcm=30, remainder=1

Ans is (A)

Note on negative numbers
I am not entirely sure what the doubt above is, but atleast in this question we are given 2<m<p, so neither can be negative.
In general when talking about division using negative numbers, there can be two values of r, and we have to decide on definition before hand, so I dont think this something the GMAT can test on. Eg, consider division of 10 by -7. 10 = 1*(-7) + (-3) & 10=2*(-7) + (4). Both -3 and 4 are well defined remainders as both satisfy the condition 0<=Abs(remainder)<Abs(divisor). I have never seen GMAT questions deal with negative divisions & remainders because of this ambiguity.

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Last edited by shrouded1 on 20 Sep 2010, 01:40, edited 1 time in total.
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Re: DS:GCD of numbers. [#permalink] New post 20 Sep 2010, 01:34
Shrouded1 :

m=10,p=30,lcm=30, remainder=0 --> is wrong , it is given that m is not a factor of P

you may take m =6 and p =10
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Re: DS:GCD of numbers. [#permalink] New post 20 Sep 2010, 01:39
gurpreetsingh wrote:
Shrouded1 :

m=10,p=30,lcm=30, remainder=0 --> is wrong , it is given that m is not a factor of P

you may take m =6 and p =10


Oops, you are right, I should take m=5, p=6, then remainder=1. Still works for the question

Editing my solution
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Re: DS:GCD of numbers. [#permalink] New post 20 Sep 2010, 03:07
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ravitejapandiri wrote:
A small doubt!Are the concepts GCD,LCM applicable to negative numbers?If so,please explain and tel me why -ve numbers are not considered in the Fact1 of the following problem?

--------------------------------------------------------------
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?

(1) the greatest common factor of m and p is 2
(2) the least common multiple of m and p is 30
--------------------------------------------------------------

I understood that Fact2 is not sufficient but am not sure about Fact1.If negative numbers are taken,then,for example, GCD(-8,6) is 2 and gives new remainder.Am I going wrong somewhere?


SOME NOTES:

1. GCD and LCM
The greatest common divisor (GCD), of two or more non-zero integers, is the largest positive integer that divides the numbers without a remainder.

So GCD can only be positive integer. It should be obvious as greatest factor of two integers can not be negative. For example if -3 is a factor of two integer then 3 is also a factor of these two integers.

The lowest common multiple (LCM), of two integers \(a\) and \(b\) is the smallest positive integer that is a multiple both of \(a\) and of \(b\).

So LCM can only be positive integer. It's also quite obvious as if we don not limit LCM to positive integer then LCM won't make sense any more. For example what would be the lowest common multiple of 2 and 3 if LCM could be negative? There is no answer to this question.

2. DIVISIBILITY QUESTIONS ON GMAT

EVERY GMAT divisibility question will tell you in advance that any unknowns represent positive integers.

3. REMAINDER

GMAT Prep definition of the remainder:
If \(a\) and \(d\) are positive integers, there exists unique integers \(q\) and \(r\), such that \(a = qd + r\) and \(0\leq{r}<d\). \(q\) is called a quotient and \(r\) is called a remainder.

Moreover many GMAT books say factor is a "positive divisor", \(d>0\).

I've never seen GMAT question asking the remainder when dividend (\(a\)) is negative, but if we'll cancel this restriction (and consider \(dividend=a<0\)), but leave the other restriction (\(0\leq{r}<d\)), then division of negative integer by positive integer could be calculated as follow:

\(-8\) divided by \(6\) will result: \(0\leq{r}<d\), \(a=qd + r\) --> \(0\leq{r}<6\), \(-8=(-2)*6+4\). Hence \(remainder=r=4\).

TO SUMMARIZE, DON'T WORRY ABOUT NEGATIVE DIVIDENDS, DIVISORS OR REMAINDERS ON GMAT.


BACK TO THE ORIGINAL QUESTION:

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?

Given: \(2<m<p\) and \(\frac{p}{m}\neq{integer}\), which means that \(r>0\). Note here that as \(0<2<m<p\) then your example of -8 and 6 is not valid as both \(m\) and \(p\) are positive.
Question: \(r=?\)

(1) the greatest common factor of m and p is 2 --> both \(p\) and \(m\) are even (as both have 2 as a factor) --> even divided by even can give only even remainder (0, 2, 4, ...), since remainder is not zero (as \(\frac{p}{m}\neq{integer}\)), then remainder must be more than 1: 2, 4, ... Sufficient.

(2) the least common multiple of m and p is 30 --> if \(m=5\) and \(p=6\), remainder=1=1, answer to the question would be NO. BUT if \(m=10\) and \(p=15\) remainder=5>1 answer to the question would be YES. Two different answers. Not sufficient.

Answer: A.

Hope it helps.

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Re: DS:GCD of numbers. [#permalink] New post 20 Sep 2010, 03:08
Already posted:
new-set-of-good-ds-85441.html

Also, posted in the wrong forum. This belongs in the quant forum.

Please search in the future before posting.

Locked.
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Re: DS:GCD of numbers. [#permalink] New post 20 Sep 2010, 03:17
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Re: DS:GCD of numbers. [#permalink] New post 20 Sep 2010, 04:34
Bunuel wrote:
Moved to the proper subforum, so now it's unlocked for further discussion.


booooo you stole my thunder :)
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Re: DS:GCD of numbers. [#permalink] New post 20 Sep 2010, 18:05
When evaluating statement (2) LCD = 30, is there a way to get the conclusion of insufficient by looking at the prime factors of 30 rather than sub-ing in numbers?
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Re: DS:GCD of numbers. [#permalink] New post 20 Sep 2010, 19:13
FQ wrote:
When evaluating statement (2) LCD = 30, is there a way to get the conclusion of insufficient by looking at the prime factors of 30 rather than sub-ing in numbers?


Its one and the same thing. You are looking for 2 factors of 30, such that one is a multiple of the other OR p=km+1 and at the same time their LCM is 30.
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Re: The integers m and p are such that 2<m<p, and m is not a [#permalink] New post 06 Jun 2013, 05:08
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Re: The integers m and p are such that 2<m<p, and m is not a [#permalink] New post 06 Jun 2013, 22:17
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ravitejapandiri 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?

(1) the greatest common factor of m and p is 2
(2) the least common multiple of m and p is 30




(1) from this statement we know that both m and n are even integers, whenever we divide even integers to each other the remainder will be 0 or more than 1. However since m and n must be greater than 2 and the greatest common factor is 2, each integer contains some distinct from each other factors. For example, 4(2*2) and 6(3*2), 6(3*2) and 8(2*2*2). So the remainder will be greater than 1 in any case. Sufficient.

(2) here we can have several integers that fit into the 2nd statement's condition and have different remainders. For example, 5 and 6, their LCM is 30 the remainder is 1; and 15 and 6, their LCM is again 30 but the remainder is greater than 1. Not sufficient.
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Re: The integers m and p are such that 2<m<p, and m is not a [#permalink] New post 10 Oct 2013, 05:52
What would be a good numerical example to satisfy statement 1
m=8, p=14?
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Re: The integers m and p are such that 2<m<p, and m is not a [#permalink] New post 10 Oct 2013, 06:22
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Re: The integers m and p are such that 2<m<p, and m is not a [#permalink] New post 30 Apr 2015, 06:06
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Re: The integers m and p are such that 2<m<p, and m is not a [#permalink] New post 21 May 2015, 00:56
What if the numbers are 10 and 22
where gcf is 2
but remainder is only 1.
shouldnt it be r>1 ? when equal to and greater wont that be insufficient ?
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Re: The integers m and p are such that 2<m<p, and m is not a [#permalink] New post 21 May 2015, 02:39
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shallow9323 wrote:
What if the numbers are 10 and 22
where gcf is 2
but remainder is only 1.
shouldnt it be r>1 ? when equal to and greater wont that be insufficient ?


22 divided by 10 gives the remainder of 2, not 1.
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Re: The integers m and p are such that 2<m<p, and m is not a [#permalink] New post 21 May 2015, 03:54
THank you

i made a mistake. I kept factoring it instead of dividing it.
Re: The integers m and p are such that 2<m<p, and m is not a   [#permalink] 21 May 2015, 03:54
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