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

1. anonymous

I found x = 82 or x = 187 but what are ALL solutions?

2. ganeshie8

did you use chinese remainder thm?

3. anonymous

I think I did.

4. anonymous

It only said Theorem but didn't explicitly said "Chinese Remainder Theorem" Let me post the link

5. ganeshie8

chinese remainder thm gives you the solution $x\equiv 82\pmod{5*7*3}$

6. beginnersmind

You can divide eq1 by 3 and eq3 by 5, since they are moduli over primes.

7. anonymous
8. anonymous

Oh so I was supposed to divide the second congruence by 2?

9. ganeshie8

that is chinese remainder theorem

10. ganeshie8

it doesn't affect the final solution dividing by 2 simplifies the work, thats all

11. beginnersmind

I don't think you can divide eq2. Actually division is the wrong word.

12. anonymous

what about the solution 187 (mod 5*14*3) ??

13. ganeshie8

in mod 5*14*3, you should get "two" incongruent solutions

14. 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

15. 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} ??

16. 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)

17. beginnersmind

Since 14 is not a prime number 4x ≡ 6 (mod 14) doesn't have a unique solution for x (mod 14)

18. anonymous

I see. Thank you both :')

19. beginnersmind

So the next step is to find all x (mod 14) that solve 4x ≡ 6 (mod 14).

20. 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 !

21. anonymous

@ganeshie8 thank you!!