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Remainders

by klaud » Thu Feb 02, 2012 12:55 pm
If M and N are positive integers that have remainders of 1 and 3, respectively, when divided by 6, which of the following could NOT be a possible value of M + N?

(A) 86
(B) 52
(C) 34
(D) 28
(E) 10

A
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by GMATGuruNY » Thu Feb 02, 2012 1:36 pm
klaud wrote:If M and N are positive integers that have remainders of 1 and 3, respectively, when divided by 6, which of the following could NOT be a possible value of M + N?

(A) 86
(B) 52
(C) 34
(D) 28
(E) 10

A
M is 1 more than a multiple of 6: 6x + 1
N is 3 more than a multiple of 6: 6y + 3
Thus:
M+N = (6x + 1) + (6y + 3) = 6(x+y) + 4.
This means that M+N is 4 more than a multiple of 6.

The answer choices represent possible values of M+N.
For an answer choice to be a valid value of M+N, it must yield a multiple of 6 when 4 is subtracted.
The correct answer will NOT yield a multiple of 6 when 4 is subtracted.

(A) 86-4 = 82.
82 is not a multiple of 6.

The correct answer is A.

Every other answer choice yields a multiple of 6 when 4 is subtracted:

(B) 52-4 = 48.
(C) 34-4 = 30.
(D) 28-4 = 24.
(E) 10-4 = 6.
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by neeti2711 » Wed Mar 06, 2013 12:45 am
Shouldn't option (E) be considered the multiple of 6?
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by Anurag@Gurome » Wed Mar 06, 2013 12:49 am
neeti2711 wrote:Shouldn't option (E) be considered the multiple of 6?
Are you trying to say 10 is multiple of 6?
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by neeti2711 » Thu Mar 07, 2013 7:45 am
4 when subtracted from 10 yields 6 which is a multiple of 6.. (6*1)..this is the reasoning.. is it wrong?
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by Anurag@Gurome » Thu Mar 07, 2013 7:52 am
neeti2711 wrote:4 when subtracted from 10 yields 6 which is a multiple of 6.. (6*1)..this is the reasoning.. is it wrong?
No, that is correct.
But the question asks "...which of the following could NOT be a possible value of M + N?"
And as Mitch has shown, (M + N) will be 4 more than a multiple of 6. Hence, the correct option should NOT yield a multiple of 6 when 4 is subtracted from it.

As 10 yields a multiple of 6 when 4 is subtracted from it, 10 is not the correct answer.

Hope that helps.
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by saurav.jha » Mon Mar 11, 2013 6:14 am
Let M be 6k1 + 1
and N be 6k2 + 3
So M+N shall be 6(k1+k2) + 4 or 6K + 4
Means the number M+N shall be multiple of 6 when 4 is subtracted from M+N
only 86 is the number above 86-4=82 not a multiple of 6.
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