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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 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.
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Originally posted by LM on 06 May 2010, 11:30.
Last edited by Bunuel on 28 Mar 2012, 01:48, edited 1 time in total.
Edited the question and added the OA




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Re: DS:GCD of numbers.
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20 Sep 2010, 03:07
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 LCMThe greatest common divisor (GCD), of two or more nonzero 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 GMATEVERY GMAT divisibility question will tell you in advance that any unknowns represent positive integers. 3. REMAINDERGMAT 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. Also: Please post PS questions in the PS subforum: gmatproblemsolvingps140/Please post DS questions in the DS subforum: gmatdatasufficiencyds141/
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The integers m and p are such that 2<m<p, and m is not a
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Updated on: 06 Jun 2013, 03:29
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?
Originally posted by ravitejapandiri on 19 Sep 2010, 23:56.
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: GMAT PREP (DS)
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06 May 2010, 12:59
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.
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Updated on: 20 Sep 2010, 01:40
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 numbersI 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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Originally posted by shrouded1 on 20 Sep 2010, 01:20.
Last edited by shrouded1 on 20 Sep 2010, 01:40, edited 1 time in total.



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Re: DS:GCD of numbers.
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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.
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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.
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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 subing in numbers?



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Re: DS:GCD of numbers.
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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 subing 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
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06 Jun 2013, 22:17
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
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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
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Re: GMAT PREP (DS)
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01 Nov 2013, 10:52
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)
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02 Nov 2013, 04:24
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)
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02 Nov 2013, 11:16
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)
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02 Nov 2013, 21:08
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
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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
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Re: The integers m and p are such that 2<m<p, and m is not a
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14 May 2016, 12:17
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
 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? 1. If r=1, that means that p=m+1, so we can try and get a definite answer for p=m+1, if not, then r>1. St1.  If p=m+1, that means that GCD(M,P)=1, since this is not the case, p is does not equal to m+1, and hence r>1. Hence, St1 sufficient. St2.  LCM(P,M)=30 > means: 1. all the distinct prime numbers are: 2,3,5. 2. since this is LCM, the max power of every prime mentioned above is 1. hence 2^1,3^1,5^1.  Now since 2<m<p, we can deduce that either m or p is prime.  case 1: m= 5 ; p= 2*3 > p=m+1  case 2: m=3; p=2*5=10 > p is not equal to m+1. hence St2 is not sufficient. Final answer: A.



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Re: The integers m and p are such that 2<m<p, and m is not a
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06 Jun 2018, 20:44
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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