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?
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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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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 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 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:
gmat-problem-solving-ps-140/Please post DS questions in the DS subforum:
gmat-data-sufficiency-ds-141/ _________________