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PS problem......ASSISTANT!

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by smclean23 » Sun Jul 06, 2008 3:10 pm
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
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by sibbineni » Sun Jul 06, 2008 3:34 pm
IMO B
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by kanak » Mon Jul 07, 2008 1:44 am
E is the ans.

Possible values:

M = { 7, 13, 19, 25, ..... }
N = { 9, 15, 21, ...... }

Minimum value of M + N = 15.

E is ruled out.
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Re: PS problem......ASSISTANT!

by sreekar.dev » Mon Jul 07, 2008 3:34 am
The answer is A.

You can have these combinations for E.

M=1 & N=9 or M=7 & N=3

So, consider these steps to solve the problem:
M=6x+1
N=6y+3

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

The only option that does not have this format is A. (86/6 = 2)
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by classic » Mon Jul 07, 2008 3:36 am
OA is A

10 can be true for value M=1 and N=9
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by classic » Mon Jul 07, 2008 3:37 am
OA is A

10 can be true for value M=1 and N=9
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by vishubn » Mon Jul 07, 2008 6:40 am
How is 10 a possiblity ????

Can none xplain plzzz?

Cheers
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by Ian Stewart » Mon Jul 07, 2008 8:33 am
kanak wrote:E is the ans.

Possible values:

M = { 7, 13, 19, 25, ..... }
N = { 9, 15, 21, ...... }

Minimum value of M + N = 15.

E is ruled out.
This is not quite right. We know M and N are positive integers; when M is divided by 6, the remainder is 1. The smallest value M can have is thus 1 (when you divide 1 by 6, the quotient is 0 and the remainder is 1). When N is divided by 6, the remainder is 3, so N can be equal to 3. The smallest possible value of M+N is 4, not 15.

M+N can certainly be 10; let M = 1 and N = 9, or let M = 7 and N = 3.

There are good solutions above that demonstrate that 86 is the lone answer choice which cannot be equal to M+N (since M+N must have a remainder of 1+3 = 4 when divided by 6).
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by AleksandrM » Mon Jul 07, 2008 9:10 am
vishubn wrote:How is 10 a possiblity ????

Can none xplain plzzz?

Cheers
M = 6z + 1 and N = 6z + 3

M + N = 12z + 4 This means that When M + N is divided by 6, the remainder will be 4.

10 divided by 6 gives you 1 whole and a remainder of 4. Try long division and you will see it for yourself. You could also see that 10/6 is 1 4/6.
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by dalwow » Wed Jul 09, 2008 11:07 am
An easier way to think about it is you know that when adding the 2 numbers together you will have a total remainder of 4 when dividing by 6. So, just take each answer choice, subtract 4 and see if it is divisible by 6. A is the only answer that is not divisible by 6 after subtracting the remainder of 4.
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Re: PS problem......ASSISTANT!

by Stuart@KaplanGMAT » Wed Jul 09, 2008 1:19 pm
smclean23 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
Another solution is to look for the pattern in the choices.

We're looking for something related to cycles of 6, so let's use that to our advantage and play the Sesame Street favourite "one of these things is not like the others".

10... 28.... 34.... 52... 86

If we look at the gaps:

18.. 6... 18... 34

Well hey! 18, 6 and 18 are all multiples of 6, so 10, 28, 34 and 52 will all fit in the same place in the multiples/remainders of 6 cycle. 86 is at a different place in the cycle, so it MUST be the right answer to the question.

There are other ways we could look at the choices as well. 10, 28, 34 and 52 all give a remainder of 4 when divided by 6, while 86 gives a remainder of 2. Even if we're not sure why we need a remainder of 4, we can deduce that 86 is the oddball and must be correct.
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