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anonymous
 one year ago
solve
3x ≡ 1 (mod 5)
4x ≡ 6 (mod 14)
5x ≡ 11 (mod 3)
anonymous
 one year ago
solve 3x ≡ 1 (mod 5) 4x ≡ 6 (mod 14) 5x ≡ 11 (mod 3)

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I found x = 82 or x = 187 but what are ALL solutions?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3did you use chinese remainder thm?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It only said Theorem but didn't explicitly said "Chinese Remainder Theorem" Let me post the link

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3chinese remainder thm gives you the solution \[x\equiv 82\pmod{5*7*3}\]

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.1You can divide eq1 by 3 and eq3 by 5, since they are moduli over primes.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0here: https://books.google.com/books?id=eVwvvwZeBf4C&lpg=PA58&pg=PA68#v=onepage&q&f=false

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh so I was supposed to divide the second congruence by 2?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3that is chinese remainder theorem

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3it doesn't affect the final solution dividing by 2 simplifies the work, thats all

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.1I don't think you can divide eq2. Actually division is the wrong word.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what about the solution 187 (mod 5*14*3) ??

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3in mod 5*14*3, you should get "two" incongruent solutions

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3``` I found x = 82 or x = 187 but what are ALL solutions? ``` Yes, those are all the incongruent solutions in mod 5*14*3

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh so in a sense {82 + 5*7*3k, k integer} is equal to the set {82 + 5*14*3k} U {187 + 5*14*3k} ??

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.1Lets take equation 1 as an example 3x ≡ 1 (mod 5) says that if you divide 3x by 5 you get a remainder of one. We want to know x (mod 5) is. Since 5 is a prime number we can determine the remainder of x:5 uniquely. In particular x ≡ 2 (mod 5). Similarly for equation three 5x ≡ 11 (mod 3) > 5x ≡ 2 (mod 3) > x ≡ 1 (mod 3)

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.1Since 14 is not a prime number 4x ≡ 6 (mod 14) doesn't have a unique solution for x (mod 14)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I see. Thank you both :')

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.1So the next step is to find all x (mod 14) that solve 4x ≡ 6 (mod 14).

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3``` oh so in a sense {82 + 5*7*3k, k integer} is equal to the set {82 + 5*14*3k} U {187 + 5*14*3k} ?? ``` Precisely @sourwing !

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@ganeshie8 thank you!!
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