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PS 1000- Section 2 #19

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by lavanya_25 » Mon May 07, 2007 7:57 am
Hi

One more that i require help with....

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

Thanks!
Lavanya
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by mschling52 » Mon May 07, 2007 8:02 am
I think it is (A). From the question stem,

M = 6k+1, for some integer k
N = 6j+3, for some integer j

M+N = 6k+6j+4 = 6(k+j)+4, where k+j is an integer. Therefore, M+N leaves a remainder of 4 when divided by 6 since 6(k+j) is evenly divisible by 6. Test the possible solutions and all have a remainder of four when divided by 6 except 86, which has a remainder of 2.
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Junior | Next Rank: 30 Posts
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OA

by lavanya_25 » Mon May 07, 2007 9:51 am
THE OA for this question is A. tHANKS a lot for explaining it to me.

Lavanya
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by Cybermusings » Mon May 07, 2007 10:51 am
10 it is
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