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What is integer n? 1. When divided by 7, remainder is 3 2.

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What is integer n? 1. When divided by 7, remainder is 3 2. [#permalink] New post 02 Apr 2006, 09:55
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What is integer n?

1. When divided by 7, remainder is 3
2. When divided by 4, remainder is 2
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 [#permalink] New post 02 Apr 2006, 10:02
Would take C

7x + 3 = 4y+2

==>> 4y-7x = 1

PLug in values y =2 and x=1.

So the number is 10.

So IMO it is C
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Re: DS: Integer n [#permalink] New post 02 Apr 2006, 10:26
vivek123 wrote:
What is integer n?

1. When divided by 7, remainder is 3
2. When divided by 4, remainder is 2


Applying Chinese modulus theorem, we get n= 28s+10 ( s is non-negative integer)
There're plenty of values of m ---> there're infinitively many values of n satisfying the two statements --> E it is!

How to apply Chinese modulus theorem:

from 1: n= 3( mod 7) (1)
from 2: n= 2 ( mod 4) (2)

from 2: n= 4v+2 , substitute this to (1), we get: 4v+2= 3(mod 7)
---> 4v= 1( mod 7) ---> v= 2 (mod 7) ( because 2*4= 8 which is divided by 7 has a remainder of 1) --> v= 7s +2 --> n= 4v+2= 4(7s+2)+2 --> n = 28s + 10
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Re: DS: Integer n [#permalink] New post 02 Apr 2006, 10:42
laxieqv wrote:
vivek123 wrote:
What is integer n?

1. When divided by 7, remainder is 3
2. When divided by 4, remainder is 2


Applying Chinese modulus theorem, we get n= 28s+10 ( s is non-negative integer)
There're plenty of values of m ---> there're infinitively many values of n satisfying the two statements --> E it is!

How to apply Chinese modulus theorem:

from 1: n= 3( mod 7) (1)
from 2: n= 2 ( mod 4) (2)

from 2: n= 4v+2 , substitute this to (1), we get: 4v+2= 3(mod 7)
---> 4v= 1( mod 7) ---> v= 2 (mod 7) ( because 2*4= 8 which is divided by 7 has a remainder of 1) --> v= 7s +2 --> n= 4v+2= 4(7s+2)+2 --> n = 28s + 10


Yes it is E.... :?

I find another set of values

when y =9 and x=5 in my solution but your looks more elegant, laxieqv

Thanks!
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 [#permalink] New post 02 Apr 2006, 12:01
Laxieqv ... thats super clever .. i havent seen that before but can you explain where this theorem came from ? and where else it could be used ?
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 [#permalink] New post 02 Apr 2006, 20:39
NTLancer wrote:
Laxieqv ... thats super clever .. i havent seen that before but can you explain where this theorem came from ? and where else it could be used ?


you can check it out here :) http://www.gmatclub.com/phpbb/viewtopic.php?t=27093
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Re: DS: Integer n [#permalink] New post 02 Apr 2006, 20:53
E.

n = 36,66....it seems strange but I found these values under 2 :).

I am not sure on the GMAT I will figure out that "C" is the trap....unless it says that the question is from vivek :).

vivek123 wrote:
What is integer n?

1. When divided by 7, remainder is 3
2. When divided by 4, remainder is 2
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Re: DS: Integer n [#permalink] New post 02 Apr 2006, 21:18
trivikram wrote:
laxieqv wrote:
vivek123 wrote:
What is integer n?

1. When divided by 7, remainder is 3
2. When divided by 4, remainder is 2


Applying Chinese modulus theorem, we get n= 28s+10 ( s is non-negative integer)
There're plenty of values of m ---> there're infinitively many values of n satisfying the two statements --> E it is!

How to apply Chinese modulus theorem:

from 1: n= 3( mod 7) (1)
from 2: n= 2 ( mod 4) (2)

from 2: n= 4v+2 , substitute this to (1), we get: 4v+2= 3(mod 7)
---> 4v= 1( mod 7) ---> v= 2 (mod 7) ( because 2*4= 8 which is divided by 7 has a remainder of 1) --> v= 7s +2 --> n= 4v+2= 4(7s+2)+2 --> n = 28s + 10


Yes it is E.... :?

I find another set of values

when y =9 and x=5 in my solution but your looks more elegant, laxieqv

Thanks!


LAXI can u explain how you got this
---> 4v= 1( mod 7) ---> v= 2 (mod 7)
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 [#permalink] New post 03 Apr 2006, 05:20
The OA is "E" :)

Laxie, thanks for the logic buddy! :good
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 [#permalink] New post 03 Apr 2006, 10:06
That is very elegant Laxie!

For people who are a little weary of the whole sequence of the Chinese modulus theorem, I hope this will help: You don't have to memorize it. :-D

Basically you write the two conditions:
n=7a+3
n=4b+2

Therefore 4b=7a-1=8a-a-1
b=2a-(a-1)/4
We know that b is an integer so (a-1)/4 must be an integer.
We can write as a-1=4s, or a=4s+1.

Now we know
a=4s+1,
b=2(4s+1)-s=7s+2

You can solve for n here from either a or b, although you don't have to. As long as you know that when s takes different values, you'll have different values for a and b and thus different values of n. In other words you'll know that the solution is not unique and the answer would be E.
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 [#permalink] New post 04 Apr 2006, 03:23
HongHu wrote:
Basically you write the two conditions:
n=7a+3
n=4b+2

Therefore 4b=7a-1=8a-a-1
b=2a-(a-1)/4
We know that b is an integer so (a-1)/4 must be an integer.
We can write as a-1=4s, or a=4s+1.

Now we know
a=4s+1,
b=2(4s+1)-s=7s+2

You can solve for n here from either a or b, although you don't have to. As long as you know that when s takes different values, you'll have different values for a and b and thus different values of n. In other words you'll know that the solution is not unique and the answer would be E.


I'm just so impressed by you, sis :)
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 [#permalink] New post 04 Apr 2006, 03:46
The method is neat! And I must say it's the first time I've heard of it. But for people who are stuck, working out with sample numbers present a very quick approach as well.

1) n = 7q1 + 3

n could be 10
n could be 15
etc

Insufficient.

2) n = 4q2 + 1
n could be 5
n could be 9
etc

Insufficient.

Using 1) and 2)
n = 7q1 + 3
n = 4q2 + 1

7q1+3 = 4q2+1
2 = 4q2-7q1

Possible sets: (q1,q2) = (4,2) (11,6) --> means many possible values of n

Ans E
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 [#permalink] New post 04 Apr 2006, 09:16
I'm impressed by you all :-D
Thanks Hong, Laxie & Wilfred! :good
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 [#permalink] New post 04 Apr 2006, 12:33
Extension to HongHu's method:

N = 7a + 3
N = 4b + 2

=> 7a+1 is divisible by 4.
Smallest possible positive value of a satisfying above equation is a = 1.
when a = 1, N = 10

LCM of 7,4 = 28

Hence, the set of all such numbers can be written as 28k + 10.

We can extend this procedure to solve more than 2 equations also.
  [#permalink] 04 Apr 2006, 12:33
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