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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] New post 04 Jun 2007, 21:05
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D
E

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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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 [#permalink] New post 04 Jun 2007, 21:18
A.

M = 6n + 1 for all n=0,1,2......
N = 6n + 3 for all n=0,1,2......

So M+N = 6n+4

M+N-4 is divisible by 6

86-4 = 82 not divisible by 6
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 [#permalink] New post 05 Jun 2007, 01:01
The OA is A, but what about E?
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 [#permalink] New post 05 Jun 2007, 04:12
Damn iamba, what a complicated way to look at it. The problem is much easier than that.

If A/6 gives a remainder of 1 and B/6 gives a remainder of 3 then (A+B)/6 should give a remainder of 1+3 = 4

Now we just look at all the options. E is fine 10 % 6 = 4 (% just means remainder when u divide). In fact the only one that doesnt fit the bill is option A where 86%6 = 2
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 [#permalink] New post 05 Jun 2007, 08:53
bsd_lover wrote:
**** iamba, what a complicated way to look at it. The problem is much easier than that.

If A/6 gives a remainder of 1 and B/6 gives a remainder of 3 then (A+B)/6 should give a remainder of 1+3 = 4

Now we just look at all the options. E is fine 10 % 6 = 4 (% just means remainder when u divide). In fact the only one that doesnt fit the bill is option A where 86%6 = 2


It looks to me like your reasoning is the same as his, except he put it into algebraic form
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 [#permalink] New post 06 Jun 2007, 15:59
I understand that with M and N together (M+N), E could be correct. But what about this:

M % 6 = 1
N % 6 = 3
Both M and N are positive integers.

IMO: M must at least equal to 7. N must at least equal to 9. So M + N must at least equal to 16. Is this trend of thought not logical?
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 [#permalink] New post 06 Jun 2007, 20:07
Minimum value of M is 1, not 7 bcos 1%6 = 1
Similarly, Minimum value of N is 3, not 9 bcos 3%6 = 3
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 [#permalink] New post 06 Jun 2007, 21:30
Ok what am i doing wrong :

m/6 = q + r
m = 6q+ 1
n = 6q + 3

m + n = 6q+1 + 6q + 3 = 12q + 4

12q + 4 =

12q = -4+86
12q = 82 - nope
12q = -4(52)
q=4
12q = -4(34) - nope
12q =-4(28)
q = 2
12q = -4(10) - nope
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 [#permalink] New post 06 Jun 2007, 21:49
I m stuck here too
y is m+n=6q+4
why not m+n= 12q+4?
some help plz
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 [#permalink] New post 07 Jun 2007, 01:34
shahrukh wrote:
I m stuck here too
y is m+n=6q+4
why not m+n= 12q+4?
some help plz


If you must do it this way

m= 6p + 1 and n = 6q + 3 where p and q are non-negative integers

m + n = 6 (p + q) + 4

In other words , m is 4 more than a multiple of 6


I hope you weren't thinking
m= 6q + 1
n = 6q +3
n + n = 12q + 4

we do not know that m and n have the same quotient when divided by 6
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 [#permalink] New post 07 Jun 2007, 22:58
I don't understand why not E
Can anybody draw an example when Mand N divided by 6 have remainders of 1 and 3, and M+N=10?
M and N must be positive though....
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 [#permalink] New post 08 Jun 2007, 00:32
N = 7
M = 3

N/6 --> remainder = 1
M/6 --> remainder = 3

M+N = 10

M+N CAN have a value of ten and satisfying the conditions mentioned in the question. THe problem asks for a value that can NOT be for M+N.

Last edited by Mishari on 08 Jun 2007, 06:28, edited 1 time in total.
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 [#permalink] New post 08 Jun 2007, 01:34
Mishari wrote:
N = 7
M = 3

N/6 --> remainder = 1
M/3 --> remainder = 3

M+N = 10

M+N CAN have a value of ten and satisfying the conditions mentioned in the question. THe problem asks for a value that can NOT be for M+N.


The problem says: If M and N are positive integers that have remainders of 1 and 3, respectively, when divided by 6.
So M must be divided by 6, not by 3 :)
If M=3 then 3/6 does not give remainder 3
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 [#permalink] New post 08 Jun 2007, 06:15
oops sorry .. but the explanation is still valid :P

3/6 has a qoutient of zero and remainder of 3
  [#permalink] 08 Jun 2007, 06:15
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If M and N are positive integers that have remainders of 1

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