Quickest way as i believe is to try out the numbers directly...
Numbers that leave 1 and 3 as remainders on dividing by 6 are:
Remainder 1 = 1,7,13,19,25... don't stretch this as maximum sum is 86 in options
Remainder 3 = 3,9,15,21,27 ......
Try you the options with minimum i.e try getting 10,28 34,52.
As you can see 86 which is option A is answer.

Is there a logic way with the possibility of NUMBER PROPERTIES & DIVISIBILITY RULES?????

ankush123251 wrote:Quickest way as i believe is to try out the numbers directly...
Numbers that leave 1 and 3 as remainders on dividing by 6 are:
Remainder 1 = 1,7,13,19,25... don't stretch this as maximum sum is 86 in options
Remainder 3 = 3,9,15,21,27 ......
Try you the options with minimum i.e try getting 10,28 34,52.
As you can see 86 which is option A is answer.

karthikpandian19 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

M leaves a remainder of 1 when divided by 6, so M=6a+1 for some positive integer, a. Likewise, N=6b+3 for some positive integer, b.

M+N = 6a+1 + 6b+3 = 6(a+b) + 4. This tells us that M+N is 4 more than a multiple of 6.

Note that all of the choices are 4 more than a multiple of 6 except for 86.

Yes logically its like this,

M = 6x + 1
N= 6y + 3

M + N = 6(x + y) + 4.
Now check the options to fit in above format.
86 does not fit thus the answer.
Note in the above method also you try and first options thus the first method i posted is quicker.

M leaves remainder of 1, which means M-1 is divisible by 6
Similarly, N-3 is divisible by 6

Therefore, M+N-4 is divisible by 6.

Trying out the options, A fits the answer

karthikpandian19 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