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What is integer n? 1. When divided by 7, remainder is 3 2. [#permalink]
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02 Apr 2006, 10:55
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What is integer n?
1. When divided by 7, remainder is 3
2. When divided by 4, remainder is 2



VP
Joined: 28 Mar 2006
Posts: 1369

Would take C
7x + 3 = 4y+2
==>> 4y7x = 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]
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02 Apr 2006, 11:26
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 nonnegative 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]
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02 Apr 2006, 11:42
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 nonnegative 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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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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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]
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02 Apr 2006, 21:53
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]
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02 Apr 2006, 22:18
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 nonnegative 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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The OA is "E"
Laxie, thanks for the logic buddy!



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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.
Basically you write the two conditions:
n=7a+3
n=4b+2
Therefore 4b=7a1=8aa1
b=2a(a1)/4
We know that b is an integer so (a1)/4 must be an integer.
We can write as a1=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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HongHu wrote: Basically you write the two conditions: n=7a+3 n=4b+2
Therefore 4b=7a1=8aa1 b=2a(a1)/4 We know that b is an integer so (a1)/4 must be an integer. We can write as a1=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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Joined: 07 Jul 2004
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Location: Singapore

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 = 4q27q1
Possible sets: (q1,q2) = (4,2) (11,6) > means many possible values of n
Ans E



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I'm impressed by you all
Thanks Hong, Laxie & Wilfred!



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Joined: 20 Feb 2006
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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.










