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

(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?

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.

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. _________________

Re: The integers m and p are such that 2<m<p, and m is not a [#permalink]

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06 Jun 2013, 23:17

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This post received KUDOS

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.
_________________

If you found my post useful and/or interesting - you are welcome to give kudos!

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?

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.
_________________

Re: The integers m and p are such that 2<m<p, and m is not a [#permalink]

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30 Apr 2015, 07:06

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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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15 Feb 2016, 07:57

Sorry, just checking. Is statement 1 not asking if the two numbers are not consecutive (as R>1). Then the first statement is stating that the GCF is 2 (for consecutive integers this is always 1) - hence suff. For statement two, mp is 30 - this could be 15 *2 but also 6*5 which are consecutive - hence not sufficient.. Where (all) am I going wrong..

Re: The integers m and p are such that 2<m<p, and m is not a [#permalink]

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14 May 2016, 13:17

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.

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