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Manager  G
Joined: 28 Jan 2017
Posts: 143
WE: Consulting (Computer Software)
For positive integers m and n, is m^2 - n^2 divisible by 3?  [#permalink]

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Difficulty:   65% (hard)

Question Stats: 52% (02:21) correct 48% (01:53) wrong based on 23 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
VP  V
Joined: 19 Oct 2018
Posts: 1174
Location: India
Re: For positive integers m and n, is m^2 - n^2 divisible by 3?  [#permalink]

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2
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

nkhl.goyal wrote:
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
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8261
GMAT 1: 760 Q51 V42 GPA: 3.82
Re: For positive integers m and n, is m^2 - n^2 divisible by 3?  [#permalink]

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1
nkhl.goyal wrote:
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

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.

_________________ Re: For positive integers m and n, is m^2 - n^2 divisible by 3?   [#permalink] 30 Nov 2019, 00:30
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