If M and N are positive intergers that have remainders of 1

Question

If M and N are positive intergers 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

oops · Answer

m = 6k+1
n=6j+3
m+n = 6(j+k) + 4

so m+n should leave a remainder of 4 when divided by 6.
all no.s above satisfy this condition, except 86.
ans=86(a)

FN · Answer

OK lets see..

M/6=K+1
N/6=K+3 

M=6K+1; N=6J+3 Where K is an integer

M+N= 6(K+J)+4 where K is some integer...


so if (M+N)/6 must give remainder 4..

so 86/6 gives remainder 2...so A it is..

lanter1 · Answer

Just looking at the answer choices, how can it be 10?

misterJJ2u · Answer

That's what threw me off when I did it the long way. Couldn't figure out how to rule out 10.

misterJJ2u · Answer

it's from that zip file of problems...there's always someone who has that link handy. if not, PM me and I'll send the packet to you.

GMATBLACKBELT · Answer

Huh thats a neat little way to solve this problem. I did as many the long way as possible in two minutes and arrived at E b/c I couldnt see how E could be an answer.

Wierd...

oops · Answer

Take m = 1, remainder = 1 when divided by 6
(the quotient can be 0, only the divisor cannot be 0)

Then take n = 9, remainder = 3 when divided by 6

So m+n=10, remainder = 4 when divided by 6

ywilfred · Answer

M = 6q1+1
N = 3q2+3 

M+N = 6q1+3q2+4 = 3(2q1+q2)+4

Back-test:
86 <-- (86-4)/3 = integer? No. Ans = A

If M and N are positive intergers that have remainders of 1

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