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When S is divided by 5 remainder is 3, when it is divided by

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When S is divided by 5 remainder is 3, when it is divided by [#permalink] New post 26 Apr 2006, 02:32
When S is divided by 5 remainder is 3, when it is divided by 7, remainder is 4. If K+S is divisible by 35, what is the least possible value of K?
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 [#permalink] New post 26 Apr 2006, 02:50
Hallo,
S=5*Z+3
s=7*G+4
When both equations are used , i got Z=(7*G+1)/5 when G=2,Z=3 and S=18. If S+K is divisible by 35 then K=17
If it is not wrong
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 [#permalink] New post 26 Apr 2006, 03:15
S = 5A + 3
S = 7B + 4

K+S is a multiple of 35

5A+3 = 7B+4
5A-7B = 1

Only possible set A,B = 3,2

S = 18

18+K/35. SMallest value of k is 35-18 = 17
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Re: Remainder [#permalink] New post 26 Apr 2006, 03:23
getzgetzu wrote:
When S is divided by 5 remainder is 3, when it is divided by 7, remainder is 4. If K+S is divisible by 35, what is the least possible value of K?


Or by Chinese modulus theorem, we have, from what is provided,
S=35n +18 ( n is integer)
S+ K= 35n +18+ K, 35n is divisible by 35 ---> the least possible value of K= 17 so that S+ K= 35n + 35
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 [#permalink] New post 26 Apr 2006, 03:39
Dude, where do I get some background theory on this Chinese Modulus theorem?
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 [#permalink] New post 26 Apr 2006, 17:39
Yes please i'd like some info as well....
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 [#permalink] New post 26 Apr 2006, 17:54
Ans: 35 -18..


Tried finding the rite value of S that gives the remainder of 3 and 4 when divided by 5, 7 resp.

Started with 30 + 3, 25 +3 , 20 +3, 15 +3!

15+3, ie 18 worked!

Hence K = 35 -18!
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 [#permalink] New post 26 Apr 2006, 17:56
Here u go .click on the link below to master chinese remainder theorem .

:)
http://www.cut-the-knot.org/blue/chinese.shtml
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 [#permalink] New post 26 Apr 2006, 18:16
Xie Xie
Danke, Thank you.. Merci!
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 [#permalink] New post 26 Apr 2006, 19:25
Quote:
Ans: 35 -18..


Tried finding the rite value of S that gives the remainder of 3 and 4 when divided by 5, 7 resp.

Started with 30 + 3, 25 +3 , 20 +3, 15 +3!

15+3, ie 18 worked!

Hence K = 35 -18!


I solved it the same way, sometimes simple logic is the efficient way to get to the answer.
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 [#permalink] New post 27 Apr 2006, 03:31
ywilfred wrote:
S = 5A + 3
S = 7B + 4

K+S is a multiple of 35

5A+3 = 7B+4
5A-7B = 1

Only possible set A,B = 3,2

S = 18

18+K/35. SMallest value of k is 35-18 = 17


A=10,B=7 is another value
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 [#permalink] New post 27 Apr 2006, 03:35
The Chinese theorem works well with 2 facts relating to S but not with 3 facts.
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Re: Remainder [#permalink] New post 27 Apr 2006, 03:37
laxieqv wrote:
getzgetzu wrote:
When S is divided by 5 remainder is 3, when it is divided by 7, remainder is 4. If K+S is divisible by 35, what is the least possible value of K?


Or by Chinese modulus theorem, we have, from what is provided,
S=35n +18 ( n is integer)
S+ K= 35n +18+ K, 35n is divisible by 35 ---> the least possible value of K= 17 so that S+ K= 35n + 35


Pls explain how you got the below stmt

S=35n +18 ( n is integer)
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Re: Remainder [#permalink] New post 27 Apr 2006, 03:49
trivikram wrote:
laxieqv wrote:
getzgetzu wrote:
When S is divided by 5 remainder is 3, when it is divided by 7, remainder is 4. If K+S is divisible by 35, what is the least possible value of K?


Or by Chinese modulus theorem, we have, from what is provided,
S=35n +18 ( n is integer)
S+ K= 35n +18+ K, 35n is divisible by 35 ---> the least possible value of K= 17 so that S+ K= 35n + 35


Pls explain how you got the below stmt

S=35n +18 ( n is integer)


We have:
S= 3(mod 5) or S= 4(mod 7)
S= 5m+ 3 ( m is integer) --> 5m+3= 4 ( mod 7) --> 5m= 1(mod 7) --> m= 3 ( mod 7 ) (the method is to try different remainders of 7, starting from 1) --> m= 7n + 3 ( n is integer) --> substitute m by n, we have:
S= 35n+ 18

The above method is so-called Chinese modulus theorem :) . I saw some above post have the link to this theorem.
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Re: Remainder [#permalink] New post 27 Apr 2006, 05:25
laxieqv wrote:
trivikram wrote:
laxieqv wrote:
getzgetzu wrote:
When S is divided by 5 remainder is 3, when it is divided by 7, remainder is 4. If K+S is divisible by 35, what is the least possible value of K?


Or by Chinese modulus theorem, we have, from what is provided,
S=35n +18 ( n is integer)
S+ K= 35n +18+ K, 35n is divisible by 35 ---> the least possible value of K= 17 so that S+ K= 35n + 35


Pls explain how you got the below stmt

S=35n +18 ( n is integer)


We have:
S= 3(mod 5) or S= 4(mod 7)
S= 5m+ 3 ( m is integer) --> 5m+3= 4 ( mod 7) --> 5m= 1(mod 7) --> m= 3 ( mod 7 ) (the method is to try different remainders of 7, starting from 1) --> m= 7n + 3 ( n is integer) --> substitute m by n, we have:
S= 35n+ 18

The above method is so-called Chinese modulus theorem :) . I saw some above post have the link to this theorem.


I like to add that the CRT works only if the mod values are relatively prime.

For three facts relating to S, I think it can be worked out but would require a very exhaustive method that invovles euclidean algorithm...

Here's a link, that shows how to solve for 3 facts using the CRT.

http://mathforum.org/library/drmath/view/56010.html

Last edited by ywilfred on 27 Apr 2006, 07:01, edited 1 time in total.
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 [#permalink] New post 27 Apr 2006, 06:58
Just to share a method someone taught me when I was still new in GMATClub. I'm not sure the logic behind it, but it works all the time.

S = 5A+3
S = 7B+4

Make B = 0, then 7B+4 = 4

Substitute 4 to A, and S = 5A+3 = 23. Now check....

23/5 --> R3
23/7 --> R2 --> No good, change to value, 23+5 = 28.

28/5 --> R3
28/7 --> R0 --> No good. Try 23-5 = 18

18/5 --> R3
18/7 --> R4 ---> This is the value of S we need.

You can make A = 0 if you like. Then we'll have 5A+3 = 3, and S = 7B+4 = 25.

Again, we need to do a little trial and error, but the required value is seldom far off.

25/5 --> R0 ---> That's no good. Try 25-7 = 18 and we'll have the required value of S to proceed with this question.
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 [#permalink] New post 27 Apr 2006, 07:44
I got an ans of 17 assuming K is always positive.
  [#permalink] 27 Apr 2006, 07:44
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