Question

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Answer 1

oh.. nevermind, ive got it!

Answer 2

could you pls share the solution

Answer 3

ok, let me type it up

Answer 4

.

Answer 5

take ur time :)

Answer 6

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

Answer 7

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 \(r\equiv 1\pmod{2\cdot 3}\)

Answer 8

so it is the only solution right?

Answer 9

Yep! there are no other solutions

Answer 10

ok thank!

Answer 11

np:)