anonymous one year ago ***

1. anonymous

oh.. nevermind, ive got it!

2. ganeshie8

could you pls share the solution

3. anonymous

ok, let me type it up

4. IrishBoy123

.

5. ganeshie8

take ur time :)

6. anonymous

But I'm not sure if thats the only solution... can someone help me check pls?

7. ganeshie8

That 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 $$r\equiv 1\pmod{2\cdot 3}$$

8. anonymous

so it is the only solution right?

9. ganeshie8

Yep! there are no other solutions

10. anonymous

ok thank!

11. ganeshie8

np:)