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Re: The sum of two positive integers, m and n, is a multiple of [#permalink]
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The sum of two positive integers, m and n, is a multiple of 3. Is n divisible by 3?
(1) When (m + 2n) is divided by 3 the remainder is 2.
(2) When (2m + n) is divided by 3 the remainder is 1.

Nice question.
m+n = 3x (where x is a quotient & this equation means that the m+n is a multiple of 3) ------Eq1

Statement 1- m+2n = 3y + 2
(m+n) + n = 3y +2
3x +n = 3y +2
n-2 = 3(y-x)
It means that n-2 is some multiple of 3 i.e. n is not a multiple of 3
Sufficient

Statement 2- 2m+n = 3y + 1
(m+n) + m = 3y +1
3x +m = 3y +1
m-1 = 3(y-x)
It means that m-1 is some multiple of 3 i.e. m is not a multiple of 3.
If m+n is a multiple of 3 but m is not a multiple of 3, then n must also not be a multiple of 3
Sufficient

Answer D

Hope this helps
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Re: The sum of two positive integers m and n is a multiple of 3.... [#permalink]
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Hi kdatt1991,

This DS question is actually built around a rare Number Property rule (know the rule will make solving this problem a lot easier); if you don't know the rule, then you can still answer the question by TESTing VALUES.

We're told that M and N are POSITIVE INTEGERS and that (M+N) is a multiple of 3.

That last 'restriction' is really important (and it's the Number Property rule) that I mentioned earlier:

Since both variables are POSITIVE INTEGERS, there are only 2 ways for (M+N) to be a multiple of 3:
1) If they're BOTH multiples of 3, then (M+N) will be a multiple of 3 (e.g. 3+3=6, 3+6=9, 12+15=27, etc.)
2) If one is NOT a multiple of 3, then the other MUST ALSO NOT be a multiple of 3 (e.g. 1+2=3, 5+4=9, 11+1=12, etc.)

The question asks if N is a multiple of 3. This is a YES/NO question.

Fact 1: (M+2N)/3 has a remainder of 2

IF....
M and N were both multiples of 3, then (M+2N)/3 would have a remainder of 0.

Since we're told that the remainder is 2, that means M and N are NOT multiples of 3, so the answer to the question is ALWAYS NO.
Fact 1 is SUFFICIENT

Fact 2: (2M+N)/3 has a remainder of 1

This is essentially the same issue we dealt with in Fact 1: IF...M and N were both multiples of 3, then (2M+N)/3 would have a remainder of 0. Since it has a remainder of 1, then M and N are NOT multiples of 3 and the answer to the question is ALWAYS NO.
Fact 2 is SUFFICIENT

Final Answer:

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Re: The sum of two positive integers, m and n, is a multiple of [#permalink]
Bunuel wrote:
Bunuel wrote:
The sum of two positive integers, m and n, is a multiple of 3. Is n divisible by 3?

Given: m + n = {multiple of 3}.

(1) When (m + 2n) is divided by 3 the remainder is 2:

m + 2n = {not a multiple of 3}
(m + n) + n = {not a multiple of 3}
{multiple of 3} + n = {not a multiple of 3}.

Thus n is not a multiple of 3. Sufficient.

(2) When (2m + n) is divided by 3 the remainder is 1:

2m + n = {not a multiple of 3}
(m + n) + m = {not a multiple of 3}
{multiple of 3} + m = {not a multiple of 3}.

Thus m is not a multiple of 3. Since m is NOT a multiple of 3 and m + n IS a multiple of 3, then n cannot be a multiple of 3. Sufficient.

Answer: D.

Hope it's clear.


GENERALLY:
If integers \(a\) and \(b\) are both multiples of some integer \(k>1\) (divisible by \(k\)), then their sum and difference will also be a multiple of \(k\) (divisible by \(k\)):
Example: \(a=6\) and \(b=9\), both divisible by 3 ---> \(a+b=15\) and \(a-b=-3\), again both divisible by 3.

If out of integers \(a\) and \(b\) one is a multiple of some integer \(k>1\) and another is not, then their sum and difference will NOT be a multiple of \(k\) (divisible by \(k\)):
Example: \(a=6\), divisible by 3 and \(b=5\), not divisible by 3 ---> \(a+b=11\) and \(a-b=1\), neither is divisible by 3.

If integers \(a\) and \(b\) both are NOT multiples of some integer \(k>1\) (divisible by \(k\)), then their sum and difference may or may not be a multiple of \(k\) (divisible by \(k\)):
Example: \(a=5\) and \(b=4\), neither is divisible by 3 ---> \(a+b=9\), is divisible by 3 and \(a-b=1\), is not divisible by 3;
OR: \(a=6\) and \(b=3\), neither is divisible by 5 ---> \(a+b=9\) and \(a-b=3\), neither is divisible by 5;
OR: \(a=2\) and \(b=2\), neither is divisible by 4 ---> \(a+b=4\) and \(a-b=0\), both are divisible by 4.

Hope it helps.


Sure, this really helps! The generic rules make it easier to understand.
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Re: The sum of two positive integers, m and n, is a multiple of [#permalink]
sps1604 wrote:
The sum of two positive integers, m and n, is a multiple of 3. Is n divisible by 3?

(1) When (m + 2n) is divided by 3 the remainder is 2.
(2) When (2m + n) is divided by 3 the remainder is 1.


ALTERNATE SOLUTION:

1) m+2n = 2, 5, 8, 11, 14....

Let's assume 3 divides n (let n = 3). Then m = -4, -1, 2, 5, 8, 11, 14...

m+n = -1, 2, 5, 8, 11, 14..., all of which are NOT divisible by 3. Thus we have reached a contradiction, and 1) is enough to tell us that n is NOT divisible by 3. SUFFICIENT.

2) 2m+n = 1, 4, 7, 10, 13, ....

Again, assume 3 divides n (let n = 3). Then 2m = -2, 1, 4, 7, 10, 13, .... Since m has to be an integer, m = -1, 2, 5, 8, 11, ...

m+n = 2, 5, 8, 11, 14, ... all of which are NOT divisible by 3. Thus, we have reached a contradiction and 2) is enough to tell us that n is NOT divisible by 3. SUFFICIENT.

Answer: D
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Re: The sum of two positive integers, m and n, is a multiple of [#permalink]
sps1604 wrote:
The sum of two positive integers, m and n, is a multiple of 3. Is n divisible by 3?

(1) When (m + 2n) is divided by 3 the remainder is 2.
(2) When (2m + n) is divided by 3 the remainder is 1.


Given: The sum of two positive integers, m and n, is a multiple of 3.
m+n = 3k
m+n= 0mod3

Asked: Is n divisible by 3?
Is n = 0mod3?

(1) When (m + 2n) is divided by 3 the remainder is 2.
m + 2n = 3k1 + 2 where k is an integer
3k + n = 3k1 + 2
n= 3(k1-k) + 2
n = 2mod3 \(\neq\) 0mod3
SUFFICIENT

(2) When (2m + n) is divided by 3 the remainder is 1.
2m + n = 3k2 + 1
m + 3k = 3k2 + 1
m = 3(k2-k) + 1
m = 1mod3 \(\neq\) 0mod3
SUFFICIENT

IMO D
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The sum of two positive integers, m and n, is a multiple of [#permalink]
Any number can be written in the form 3a or 3a + 1 or 3a + 2. Now as per the question m + n is a multiple of 3, this means :
Case 1: m, n both are multiples of 3
Case 2: m is of the form 3a+1 and n is of the form 3b+2
Case 3: reverse of case 2.

Now Statement 1: m+2n gives remainder 2 when divided by 3. This is possible only for case 2. Hence n is not a multiple of 3, hence sufficient.
Statement 2: 2m + n gives remainder 1 when divided by 3. This is again possible only for case 2. Hence again sufficient.
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Re: The sum of two positive integers, m and n, is a multiple of [#permalink]
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