M & N

[rohangupta83]

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

I picked E but the answer given is A.

I don't get it as I chose M = 7 and N = 9 (the smallest M and N that are possible) and their sum is surely more than 10 making it the one that could not be the value of M+N.

Please let me know what's wrong with my reasoning

[sudhir3127]

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

I picked E but the answer given is A.

I don't get it as I chose M = 7 and N = 9 (the smallest M and N that are possible) and their sum is surely more than 10 making it the one that could not be the value of M+N.

Please let me know what's wrong with my reasoning

We know from the question that
M= 6x+1 ..........................................1
N = 6y+3...........................................2

if u add 1 and 2 ... u will get an equation which is a multiple of 6 plus 4

everything other than A satisifes this ...

hope this is clear.. do let me know if u have any doubts...

[rohangupta83]

ahh.. got it!

So, I had to set this equation and make a series.

I'll the above approach in mind now

Thanks Sudhir

[jnellaz]

My apologies Sudhir, could you explain your thinking a little further? I am not following fully.

Thanks.

[rohangupta83]

My apologies Sudhir, could you explain your thinking a little further? I am not following fully.

Thanks.

basically Sudhir's solution is:

By adding the two equations, we get:

M+N = 6x+6y+4

M+N = 6(x+y) +4

if x+y = 1
M+N = 10

x+y = 2
M+N = 16

X+Y = 3
M+N = 22

and so on... so the series is 10,16,22,28,34,40,46,52,58,64,70,76,82,88...

So, 86 is not present in the series so it cannot be the sum of M+N

[vishubn]

great Solution sudhir

[logitech]

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

I picked E but the answer given is A.

I don't get it as I chose M = 7 and N = 9 (the smallest M and N that are possible) and their sum is surely more than 10 making it the one that could not be the value of M+N.

Please let me know what's wrong with my reasoning

We know from the question that
M= 6x+1 ..........................................1
N = 6y+3...........................................2

if u add 1 and 2 ... u will get an equation which is a multiple of 6 plus 4

everything other than A satisifes this ...

hope this is clear.. do let me know if u have any doubts...

Yeap,

M+N = 6(x+y)+4

In other words, if you subtract 4 from the answer choices, they need to be divided by 6

[iamcste]

Using options is the best solution here

From stem

M+N=6(Q1+Q2)+4----

Q1 and Q2 are quotients

This means Q1+Q2 should exactly divide M+N

From options, we know M+N, and hence from above equation we can get (Q1 +Q2)

If Q1+Q2=0, Thats possible value of M+N

If Q1+Q2 has some remainder, Thats not possible value of M+N

Considering our options, only the first one, we get remainder , hence thats not possible value of M+N