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28 Nov 2019, 01:35
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
53% (02:06) correct
47%
(01:55)
wrong
based on 177
sessions
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For positive integers m and n, is m^2 - n^2 divisible by 3?
(1) m - n is divisible by 3
(2) When m^2 + n^2 is divided by 3, the remainder is 2
(1) m - n is divisible by 3
(2) When m^2 + n^2 is divided by 3, the remainder is 2
28 Nov 2019, 05:34
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Statement 1-
m-n is divisible by 3
\(m^2-n^2 = (m+n)(m-n)\)
Hence, \(m^2-n^2\) must be divisible by 3
Sufficient
Statement 2-
When\( m^2 + n^2\) is divided by 3, the remainder is 2
Square of any positive integer leaves 0 or 1 as a remainder, when divided by 3.
hence, \(m^2= 3a+1\) and \(n^2= 3b+1\)
\(m^2-n^2\)= 3a+1-3b-1= 3(a-b)
Sufficient
{proof of Square of any positive integer leaves 0 or 1 as a remainder, when divided by 3.}
We can write any positive integer as 3k, 3k+1 or 3k+2
Case 1-
N=3k; \(N^2=9k^2 \)
N^2 will leave 0 remainder, when divided by 3
Case 2-
N=3k+1;
N^2=\( (3K+1)^2\)
= \(9K^2+6K+1\)
= \(3(3k^2+2k) \)+1
Case 3-
N=3k+2;
\(N^2= (3K+2)^2\)
=\( 9K^2+6K+4\)
=\(9K^2+6K+3+1\)
= \(3(3k^2+2k+1)\) +1
m-n is divisible by 3
\(m^2-n^2 = (m+n)(m-n)\)
Hence, \(m^2-n^2\) must be divisible by 3
Sufficient
Statement 2-
When\( m^2 + n^2\) is divided by 3, the remainder is 2
Square of any positive integer leaves 0 or 1 as a remainder, when divided by 3.
hence, \(m^2= 3a+1\) and \(n^2= 3b+1\)
\(m^2-n^2\)= 3a+1-3b-1= 3(a-b)
Sufficient
{proof of Square of any positive integer leaves 0 or 1 as a remainder, when divided by 3.}
We can write any positive integer as 3k, 3k+1 or 3k+2
Case 1-
N=3k; \(N^2=9k^2 \)
N^2 will leave 0 remainder, when divided by 3
Case 2-
N=3k+1;
N^2=\( (3K+1)^2\)
= \(9K^2+6K+1\)
= \(3(3k^2+2k) \)+1
Case 3-
N=3k+2;
\(N^2= (3K+2)^2\)
=\( 9K^2+6K+4\)
=\(9K^2+6K+3+1\)
= \(3(3k^2+2k+1)\) +1
nkhl.goyal
General Discussion
30 Nov 2019, 00:30
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nkhl.goyal
Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Visit https://www.mathrevolution.com/gmat/lesson for details.
The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We should simplify conditions if necessary.
The questions asks if \(m^2-n^2 = (m-n)(m+n)\) is divisible by \(3\).
Since \(m-n\) is divisible by \(3\) from condition 1), \(m^2-n^2 = (m-n)(m+n)\) is divisible by \(3\) and condition 1) is sufficient.
Condition 2)
If \(k\) is not divisible by \(3\), \(k^2\) has a remainder \(1\) when it is divided by \(3\) for the followings.
If \(k = 3a + 1\), then \(k^2=(3a+1)^2 = 9a^2+6a+1 = 3(3a^2+2a)+1\).
If \(k = 3a + 2\), then \(k^2=(3a+2)^2 = 9a^2+12a+4 = 3(3a^2+4a+1)+1\).
Thus, the condition "\(m^2+n^2\) has remainder \(2\) when it is divided by \(3\)" means that \(m\) and \(n\) are not divisible by \(3\).
Then we have four cases.
Case 1: \(m=3b+1, n=3c+1\)
\((m-n)(m+n)=(3b+1-3c-1)(3b+1+3c+1)=(3b-3c)(3b+3c+2)=3(b-c)(3b+3c+2)\) is divisible by \(3\).
Case 2: \(m=3b+1, n=3c+2\)
\((m-n)(m+n)=(3b+1-3c-2)(3b+1+3c+2)=(3b-3c-1)(3b+3c+3)=3(3b-3c-1)(b+c+1)\) is divisible by \(3\).
Case 3: \(m=3b+2, n=3c+1\)
\((m-n)(m+n)=(3b+2-3c-1)(3b+2+3c+1)=(3b-3c+1)(3b+3c+3)=3(3b-3c+1)(b+c+1)\) is divisible by \(3\).
Case 4: \(m=3b+2, n=3c+2\)
\((m-n)(m+n)=(3b+2-3c-2)(3b+2+3c+2)=(3b-3c)(3b+3c+3)=3(b-c)(3b+3c+4)\) is divisible by \(3\).
Since condition 2) yields a unique solution, it is sufficient.
Therefore, D is the answer.
