A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 one year ago
***
anonymous
 one year ago
***

This Question is Closed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh.. nevermind, ive got it!

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0could you pls share the solution

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok, let me type it up

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0But I'm not sure if thats the only solution... can someone help me check pls?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0That looks good! Just for the sake of an alternative : \[\begin{align*} x &\equiv r \pmod{6}, \\ x &\equiv 9 \pmod{20}, \\ x &\equiv 4 \pmod{45} \end{align*}\] From first and second congruence, notice that \(\gcd(6,20)=2\), so it must be the case that \[r\equiv 9\equiv 1\pmod{2}\tag{a}\]. From first and last congruences, notice that \(\gcd(6,45)=3\), so it must be the case that \[r\equiv 4\equiv 1\pmod{3}\tag{b}\] From \((a),~(b)\) it follows \(r\equiv 1\pmod{2\cdot 3}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so it is the only solution right?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0Yep! there are no other solutions
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.