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If M and N are positive integers that have remainders of 1

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Senior Manager
Joined: 12 Mar 2007
Posts: 273
If M and N are positive integers that have remainders of 1 [#permalink]

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04 Jun 2007, 22:05
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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
Manager
Joined: 07 May 2007
Posts: 178

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04 Jun 2007, 22:18
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
Senior Manager
Joined: 12 Mar 2007
Posts: 273

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05 Jun 2007, 02:01
The OA is A, but what about E?
CEO
Joined: 17 May 2007
Posts: 2950

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05 Jun 2007, 05:12
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
Manager
Joined: 12 Apr 2007
Posts: 168

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05 Jun 2007, 09:53
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
Senior Manager
Joined: 12 Mar 2007
Posts: 273

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06 Jun 2007, 16:59
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?
Manager
Joined: 07 May 2007
Posts: 178

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06 Jun 2007, 21:07
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
Director
Joined: 10 Feb 2006
Posts: 657

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06 Jun 2007, 22:30
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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Manager
Joined: 17 Oct 2006
Posts: 52

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06 Jun 2007, 22:49
I m stuck here too
y is m+n=6q+4
why not m+n= 12q+4?
some help plz
GMAT Instructor
Joined: 04 Jul 2006
Posts: 1262

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07 Jun 2007, 02:34
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
Senior Manager
Joined: 04 Mar 2007
Posts: 434

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07 Jun 2007, 23:58
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....
Director
Joined: 30 Nov 2006
Posts: 591
Location: Kuwait

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08 Jun 2007, 01:32
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, 07:28, edited 1 time in total.
Senior Manager
Joined: 04 Mar 2007
Posts: 434

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08 Jun 2007, 02:34
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
Director
Joined: 30 Nov 2006
Posts: 591
Location: Kuwait

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08 Jun 2007, 07:15
oops sorry .. but the explanation is still valid

3/6 has a qoutient of zero and remainder of 3
08 Jun 2007, 07:15
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