anonymous
  • anonymous
solve 3x ≡ 1 (mod 5) 4x ≡ 6 (mod 14) 5x ≡ 11 (mod 3)
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
I found x = 82 or x = 187 but what are ALL solutions?
ganeshie8
  • ganeshie8
did you use chinese remainder thm?
anonymous
  • anonymous
I think I did.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
It only said Theorem but didn't explicitly said "Chinese Remainder Theorem" Let me post the link
ganeshie8
  • ganeshie8
chinese remainder thm gives you the solution \[x\equiv 82\pmod{5*7*3}\]
beginnersmind
  • beginnersmind
You can divide eq1 by 3 and eq3 by 5, since they are moduli over primes.
anonymous
  • anonymous
here: https://books.google.com/books?id=eVwvvwZeBf4C&lpg=PA58&pg=PA68#v=onepage&q&f=false
anonymous
  • anonymous
Oh so I was supposed to divide the second congruence by 2?
ganeshie8
  • ganeshie8
that is chinese remainder theorem
ganeshie8
  • ganeshie8
it doesn't affect the final solution dividing by 2 simplifies the work, thats all
beginnersmind
  • beginnersmind
I don't think you can divide eq2. Actually division is the wrong word.
anonymous
  • anonymous
what about the solution 187 (mod 5*14*3) ??
ganeshie8
  • ganeshie8
in mod 5*14*3, you should get "two" incongruent solutions
ganeshie8
  • ganeshie8
``` 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
  • anonymous
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} ??
beginnersmind
  • beginnersmind
Lets 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
  • beginnersmind
Since 14 is not a prime number 4x ≡ 6 (mod 14) doesn't have a unique solution for x (mod 14)
anonymous
  • anonymous
I see. Thank you both :')
beginnersmind
  • beginnersmind
So the next step is to find all x (mod 14) that solve 4x ≡ 6 (mod 14).
ganeshie8
  • ganeshie8
``` 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
  • anonymous
@ganeshie8 thank you!!

Looking for something else?

Not the answer you are looking for? Search for more explanations.