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The sum of two positive integers, m and n, is a multiple of [#permalink]
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13 Jun 2013, 05:57
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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.
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Re: The sum of two positive integers, m and n, is a multiple of [#permalink]
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13 Jun 2013, 06:07




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Re: The sum of two positive integers, m and n, is a multiple of [#permalink]
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13 Jun 2013, 06:09
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 n2 = 3(yx) It means that n2 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 m1 = 3(yx) It means that m1 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 [#permalink]
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13 Jun 2013, 06:12
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 \(ab=3\), 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 \(ab=1\), 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 \(ab=1\), is not divisible by 3; OR: \(a=6\) and \(b=3\), neither is divisible by 5 > \(a+b=9\) and \(ab=3\), neither is divisible by 5; OR: \(a=2\) and \(b=2\), neither is divisible by 4 > \(a+b=4\) and \(ab=0\), both are divisible by 4. Hope it helps.
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Re: The sum of two positive integers, m and n, is a multiple of [#permalink]
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13 Jun 2013, 07:27
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 \(ab=3\), 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 \(ab=1\), 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 \(ab=1\), is not divisible by 3; OR: \(a=6\) and \(b=3\), neither is divisible by 5 > \(a+b=9\) and \(ab=3\), neither is divisible by 5; OR: \(a=2\) and \(b=2\), neither is divisible by 4 > \(a+b=4\) and \(ab=0\), 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]
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30 May 2014, 07:42
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 3.... [#permalink]
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03 Feb 2015, 21:18
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: GMAT assassins aren't born, they're made, Rich
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