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Re: GMAT PREP (DS) [#permalink]
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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?

Given: \(2<m<p\) and \(\frac{p}{m}\neq{integer}\). \(p=xm+r\). q: \(r=?\)

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

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

Answer: A.
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Re: DS:GCD of numbers. [#permalink]
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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.

Originally posted by shrouded1 on 20 Sep 2010, 02:20.
Last edited by shrouded1 on 20 Sep 2010, 02:40, edited 1 time in total.
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Re: DS:GCD of numbers. [#permalink]
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]
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]
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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]
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]
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fozzzy wrote:
What would be a good numerical example to satisfy statement 1
m=8, p=14?


Yes. Or:
m=4, and p=6.
m=4, and p=10.
m=4, and p=14.
m=4, and p=18.
...
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Re: GMAT PREP (DS) [#permalink]
Bunuel 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?

Given: \(2<m<p\) and \(\frac{p}{m}\neq{integer}\). \(p=xm+r\). q: \(r=?\)

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

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

Answer: A.


I just have one doubt which seems going against my basic concepts when 15 is divided by 10, it has same result as when 3 is divided by 2, and when you divide 15 by 10, you do not take it as 3 divided by 2. The remainders of both the fractions should 1, but you are taking it otherwise.

Pl explain this part.
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Re: GMAT PREP (DS) [#permalink]
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ankit41 wrote:
Bunuel 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?

Given: \(2<m<p\) and \(\frac{p}{m}\neq{integer}\). \(p=xm+r\). q: \(r=?\)

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

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

Answer: A.


I just have one doubt which seems going against my basic concepts when 15 is divided by 10, it has same result as when 3 is divided by 2, and when you divide 15 by 10, you do not take it as 3 divided by 2. The remainders of both the fractions should 1, but you are taking it otherwise.

Pl explain this part.


Not sure I understand what you mean. Can you please elaborate? Thank you.

Both 15/10 and 3/2 is 1.5 but the remainder when 15 is divided by 10 is 5 and the remainders when 3 is divided by 2 is 1.
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Re: GMAT PREP (DS) [#permalink]
Bunuel wrote:
ankit41 wrote:
Bunuel 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?

Given: \(2<m<p\) and \(\frac{p}{m}\neq{integer}\). \(p=xm+r\). q: \(r=?\)

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

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

Answer: A.


I just have one doubt which seems going against my basic concepts when 15 is divided by 10, it has same result as when 3 is divided by 2, and when you divide 15 by 10, you do not take it as 3 divided by 2. The remainders of both the fractions should 1, but you are taking it otherwise.

Pl explain this part.


Not sure I understand what you mean. Can you please elaborate? Thank you.

Both 15/10 and 3/2 is 1.5 but the remainder when 15 is divided by 10 is 5 and the remainders when 3 is divided by 2 is 1.



Why is the remainder of 15/10 is 5 and not 1? we know that 15/10 is same as 3/2. What I am doing while calculating remainders, I simplify the fraction if it can be further simplified like in the case of 15/10, I simplified the fraction to 3/2 and then I will find the remainder, it comes out to be 1 but If I donot simplify the fraction the remainder comes out to be 5.
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Re: GMAT PREP (DS) [#permalink]
ankit41 wrote:
Why is the remainder of 15/10 is 5 and not 1? we know that 15/10 is same as 3/2. What I am doing while calculating remainders, I simplify the fraction if it can be further simplified like in the case of 15/10, I simplified the fraction to 3/2 and then I will find the remainder, it comes out to be 1 but If I donot simplify the fraction the remainder comes out to be 5.


Hmm. Well what you are doing is not entirely correct. When you simplify the given fractions, keep in hand the common factor which factored out from both the Numerator and the Denominator. For eg, in \(\frac{15}{10}\) we factored out 5 from both the places. Thus, the remainder of the simplified fraction, as you correctly said is 1. Just multiply this back with the common factor, i.e. in this case 5. And hence, the final remainder is 5.

You cannot just cancel out factors and expect the same remainder.

For eg, if your logic were true, we would have Remainder of \(\frac{16}{6}\) = Remainder of \(\frac{8}{3}\) which is definitely not the case.
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Re: The integers m and p are such that 2<m<p, and m is not a [#permalink]
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]
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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]
from 1st hint what if i take m=6 and p=10 then remainder comes out to be 4 .
then answer shouldn't be (E) ?

Can someone please explain where i am going wrong ?
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Re: The integers m and p are such that 2<m<p, and m is not a [#permalink]
chauhanvibhu wrote:
from 1st hint what if i take m=6 and p=10 then remainder comes out to be 4 .
then answer shouldn't be (E) ?

Can someone please explain where i am going wrong ?


YES, the remainder comes out to be 4, But the question is asking whether the remainder is > 1 or not? 4 is > 1, so answer to the question asked is YES. Sufficient to answer still. No matter which example you take, you have to take both even values such that one is not a factor of other, and remainder will thus be either 2 or 4 or 6... always greater than 1.
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Re: The integers m and p are such that 2<m<p, and m is not a [#permalink]
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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?

(1) The greatest common factor of m and p is 2.

(2) The least common multiple of m and p is 30.


m is not a factor of p.

So when p is divided by m, it will not leave remainder 0. We need to find whether the remainder "can be 1" or it "must be greater than 1".

(1) The greatest common factor of m and p is 2.

m and p are both even but p is not a multiple of m. So p will be at least 2 away from m or its multiple. e.g. if m is 4, p cannot be 8 but it can be 10, p cannot be 12 but it can be 14.
So when you divide p by m, you will get at least 2 as remainder. So remainder will be more than 1.
Sufficient.

(2) The least common multiple of m and p is 30.
30 = 2 * 3 *5
m and p could be 5 and 6 (remainder 1)
or m and p could be 6 and 10 (remainder 4)
Not sufficient.

Answer (A)
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