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1000PS--Q.19

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1000PS--Q.19

by prachich1987 » Thu Dec 16, 2010 5:18 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


The OA is A. But I think M+N cannot be 34.Also it cannot be 10.Is this Q wrong?
https://gmat1000ps.blocked/2010/01/ ... 2-q19.html
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by beat_gmat_09 » Thu Dec 16, 2010 5:33 am
6x+1=M, where x is an integer
6y+3=N, where y is an integer

M+N
=6x+1+6y+3
=6(x+y)+4


Plug option answers to get x+y, the option which gives x+y non-integer value is the answer

A)86 - x+y non-integer, 84/6 = 42/3

B)52 - x+y integer, 48/6=8

C)34 - x+y integer, 30/6=5

D)28 - x+y integer, 24/6=4

E)10 - x+y integer, 6/6=1
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by shovan85 » Thu Dec 16, 2010 5:33 am
prachich1987 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 = 6k + 1 where k is an integer
N = 6k' + 3 where k' is an integer

Thus M+N = 6(k+k')+4

We are sure that k+k' is an integer.

In order to get the answer (quick) attack the Options with the above information. Subtract 4 from each and the option which is not divisible by 6 is your desired answer (as k+k' is an integer)
Now the Options:

(A) 86: 86-4 = 82 Not divisible by 6
(B) 52: 52-4 = 48 divisible by 6
(C) 34: 34-4 = 30 divisible by 6
(D) 28: 28-4 = 24 divisible by 6
(E) 10: 10-4 = 6 divisible by 6

Thus the answer is clearly A

Let me know if something else is bugging you ;)
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by blaster » Thu Dec 16, 2010 5:44 am
M/6=x+1
N/6=s+3

M+N = x+1+s+3=x+s+4

so number must be divisible to 6 if we take 4 from it.
let's check answers

(A) 86 86-4=82 is not divisible to 6
(B) 52 52-4=48 is divisible
(C) 34 30-4=30 divisible
(D) 28 28-4=24 divisible
(E) 10 10-4=6 divisible
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by prachich1987 » Thu Dec 16, 2010 6:05 am
Thanks beat_gmat_09, shovan85,blaster.I was doing a silly mistake
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