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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