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M & N

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M & N

by rohangupta83 » Thu Nov 06, 2008 3:35 am
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
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Re: M & N

by sudhir3127 » Thu Nov 06, 2008 3:41 am
rohangupta83 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

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...
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by rohangupta83 » Thu Nov 06, 2008 3:49 am
ahh.. got it!

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

I'll the above approach in mind now

Thanks Sudhir
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by jnellaz » Fri Nov 07, 2008 11:09 am
My apologies Sudhir, could you explain your thinking a little further? I am not following fully.

Thanks.
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by rohangupta83 » Fri Nov 07, 2008 11:35 am
jnellaz wrote: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
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by vishubn » Fri Nov 07, 2008 7:30 pm
great Solution sudhir :)
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Re: M & N

by logitech » Fri Nov 07, 2008 7:37 pm
sudhir3127 wrote:
rohangupta83 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

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
LGTCH
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"DON'T LET ANYONE STEAL YOUR DREAM!"
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IMO A

by iamcste » Sat Nov 08, 2008 12:00 pm
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
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