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by bww » Sun May 06, 2007 5:11 pm
Can someone give me a sanity check? I'm stuck. How can 10 be a possible value?

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

OA is E.
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Re: prep question

by jayhawk2001 » Sun May 06, 2007 5:34 pm
bww wrote:Can someone give me a sanity check? I'm stuck. How can 10 be a possible value?

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

OA is E.
I'm getting A

M = 6x + 1
N = 6y + 3

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

A: 86-4 = 82 not divisible by 6
B: 52-4 = 48 divisible by 6
C: 30 divisible by 6
D: 24 divisible by 6
E: 6 divisible by 6

In fact if M = 1 and N = 9, we get M+N = 10 and this also satisfies
the remainder constraints.
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by kookoo4tofu » Mon May 07, 2007 7:00 am
dont forget M and N MUST be positive integers so (x+y) must be greater than 1 no matter what, thus the sum must be greater than 10
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by kookoo4tofu » Mon May 07, 2007 7:01 am
dont forget M and N MUST be positive integers so (x+y) must be greater than 1 no matter what, thus the sum must be greater than 10
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by Cybermusings » Mon May 07, 2007 10:42 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

M can belong to the set {7,13,19,25,31,37....}
N belongs to the set {9,15,21,27,33,39,45...}

thus the least value of M+N can be 7+9 = 16
Hence 10 is clearly ruled out

So E
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by f2001290 » Sat May 19, 2007 7:57 am
Cybermusings

If M = 7 and N = 3, then 10 is satisfies

I will go with A .
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by RAGS » Sat May 19, 2007 8:55 am
The ans is A
Sincex+y will leave a remainder of 4 when div by 6
only A does not
If It Is To Be It Is Up To Me
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