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
M & N
This topic has expert replies
-
- Legendary Member
- Posts: 541
- Joined: Thu May 31, 2007 6:44 pm
- Location: UK
- Thanked: 21 times
- Followed by:3 members
- GMAT Score:680
-
- Legendary Member
- Posts: 829
- Joined: Mon Jul 07, 2008 10:09 pm
- Location: INDIA
- Thanked: 84 times
- Followed by:3 members
We know from the question thatrohangupta83 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
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...
-
- Legendary Member
- Posts: 541
- Joined: Thu May 31, 2007 6:44 pm
- Location: UK
- Thanked: 21 times
- Followed by:3 members
- GMAT Score:680
ahh.. got it!
So, I had to set this equation and make a series.
I'll the above approach in mind now
Thanks Sudhir
So, I had to set this equation and make a series.
I'll the above approach in mind now
Thanks Sudhir
-
- Legendary Member
- Posts: 541
- Joined: Thu May 31, 2007 6:44 pm
- Location: UK
- Thanked: 21 times
- Followed by:3 members
- GMAT Score:680
basically Sudhir's solution is:jnellaz wrote:My apologies Sudhir, could you explain your thinking a little further? I am not following fully.
Thanks.
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
- logitech
- Legendary Member
- Posts: 2134
- Joined: Mon Oct 20, 2008 11:26 pm
- Thanked: 237 times
- Followed by:25 members
- GMAT Score:730
Yeap,sudhir3127 wrote:We know from the question thatrohangupta83 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
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...
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
---------------------
"DON'T LET ANYONE STEAL YOUR DREAM!"
---------------------
"DON'T LET ANYONE STEAL YOUR DREAM!"
-
- Legendary Member
- Posts: 940
- Joined: Tue Aug 26, 2008 3:22 am
- Thanked: 55 times
- Followed by:1 members
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
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