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any other method

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any other method

by raunekk » Fri Jun 27, 2008 1:56 am
19. 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


is there any other way to solve tis problem??

what would be the fastest way to solve such problems..??

OA=A[spoiler][/spoiler]
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by floravaze » Fri Jun 27, 2008 3:08 am
hi,
first,are you sure the answer is A.Because it seems to me like the right answer should be E.let me explain.
when M is divided by 6,remainder is 1,so let's have an equation
M/6=x + 1
When N is divided by 6,remainder is 3,so
N/6=y + 3
When we cross-multiply,we get M=6x +6 and N=6y+18
Therefore M+N=9x+6y+24.
Looking at the answer choices,we see that the most unlikely choice is 10 coz the above equation already tells us that M+N must be greater than or equal to 24.Hope this helps.
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by niraj_a » Fri Jun 27, 2008 4:34 am
i believe it is A and here's my thought process,

if remainder of M is 1, then M could be 6, and similarly N could be 9. then if 9+7 = 16 is divided 6, then remainder we get is 4.

so if we backsolve with the answer choices, any answer choice that doesn't yield a remainder 4 when divided by 6 should be our answer.

so 86/6 meets that logic so A.
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by tanviet » Fri Jun 27, 2008 6:09 am
M=6x+1
N=6y+3

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

x+y must be bigger than 2, M+N must bigger than 16

E is the right answer
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by Ian Stewart » Fri Jun 27, 2008 7:37 am
duongthang wrote:M=6x+1
N=6y+3

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

x+y must be bigger than 2, M+N must bigger than 16

E is the right answer
This starts out well, but there is no reason why x+y must be larger than 2. There is nothing wrong with having a quotient of 0 in division: when you divide 5 by 6, for example, the quotient is 0 and the remainder is 5. x and y are the quotients here, and they can certainly be equal to 0.

That said, your approach is good to begin with. We know

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

That means that M+N must be 4 larger than a multiple of 6, or in other words, that the remainder must be 4 when you divide M+N by 6. This is essentially what niraj found above by picking numbers. A is the only answer that gives a remainder different from 4 when you divide by 6.
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by tokaitalbo » Fri Jun 27, 2008 7:52 am
For me, doing this out seems the easiest, and fastest.

if the remainder is 1, then M can only be 1,7,13,19,25
if the remainder is 3, then N can only be 3,9,15,21,27

start at the bottom (smallest) and use POE

cannot be E: 1+9=10
cannot be D: 27+1=28
cannot be C: 27+7=34
cannot be B: 25+27=52.

Answer must be A.
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Wrong approach?

by yvichman » Fri Jun 27, 2008 7:58 am
Can't you just divide each of the answer choices given, by 6 to find out if they give a remainder of 4 (3+1 as stated in the question). The one that does not have a remainder of 4 would give you the correct answer.

Is this an approach that would not work all the time?
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Re: Wrong approach?

by Ian Stewart » Fri Jun 27, 2008 8:26 am
yvichman wrote:Can't you just divide each of the answer choices given, by 6 to find out if they give a remainder of 4 (3+1 as stated in the question). The one that does not have a remainder of 4 would give you the correct answer.

Is this an approach that would not work all the time?
Yes, that's perfect. My post above is identical in method- I just wanted to demonstrate why the remainder needs to be equal to 4.
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