If M and N are positive integers that have remainders of 1 : Quant Question Archive [LOCKED]
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# If M and N are positive integers that have remainders of 1

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If M and N are positive integers that have remainders of 1 [#permalink]

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16 Nov 2006, 21:10
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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

Can anyone show me an easy way of solving this...
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16 Nov 2006, 21:34
M=6n1+1
N=6n2+3

M+N = 6(n1+n2)+4=6(x)+4

So, M+N will be a multiple of 6, + 4.

Except 86, all the other values meet this criteria.

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16 Nov 2006, 22:04
M = 6n + 1
N = 6t + 3

M+n = 6(n+t) + 4 = 2(3(n+t) + 2)

A - (86/2) - 2 = 41 <-- not divisible by 3
B - (52/2) - 2 = 48 <-- divisible by 3
C - (34/2) - 2 = 15 <-- divisible by 3
D - (28/2) - 2 = 12 <-- divisible by 3
E - (10/2) - 2 = 3 <--- divisible by 3

Ans A
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16 Nov 2006, 22:28
Another approach

(m + n) / 6 = x + 4

6x + 24 = 86, x = 7

All other numbers when tested are not divisibly by 6 hence A is the answer
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16 Nov 2006, 23:29
My way:
Pick numbers

m=7
n=9

m+n=16...divide by 6....remainder is 4

Now look at the answer choices. Divide each value by 6 and look where the raimainder isn't 4.

A
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17 Nov 2006, 03:13
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= 6x+1

n= 6y+3

m+n= 6x+6y+4

86/6= 14 2/6 ( remainder is 2 not 4

17 Nov 2006, 03:13
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