Problem
Solve $f(x) = 6x^3 + 27x^2 + 17x + 20 \equiv 0 \pmod{30}$
My attempt was:
Since $30 = 2.3.5$, we then have:
$ \begin{cases} f(x) \equiv 0 \pmod{2}\\ f(x) \equiv 0 \pmod{3}\\ f(x) \equiv 0 \pmod{5}\\ \end{cases} $
By inspection, we see that: $ \begin{cases} x \equiv 0 \pmod{2}\\ x \equiv 1 \pmod{2}\\ x \equiv 2 \pmod{3}\\ x \equiv 0 \pmod{5}\\ x \equiv 1 \pmod{5}\\ \end{cases} $
Hence, there will be four cases:
Case 1 $ \begin{cases} x \equiv 0 \pmod{2}\\ x \equiv 2 \pmod{3}\\ x \equiv 1 \pmod{5}\\ \end{cases} $ Case 2
$ \begin{cases} x \equiv 0 \pmod{2}\\ x \equiv 2 \pmod{3}\\ x \equiv 0 \pmod{5}\\ \end{cases} $ Case 3 $ \begin{cases} x \equiv 1 \pmod{2}\\ x \equiv 2 \pmod{3}\\ x \equiv 1 \pmod{5}\\ \end{cases} $ Case 4 $ \begin{cases} x \equiv 1 \pmod{2}\\ x \equiv 2 \pmod{3}\\ x \equiv 0 \pmod{5}\\ \end{cases} $
Apply Chinese Remainder Theorem for the four system of equation, where $M = 30$:
$ \begin{cases} M_1 = \frac{30}{2} = 15\\ M_2 = \frac{30}{3} = 10\\ M_3 = \frac{30}{5} = 6\\ \end{cases} $ And, $ \begin{cases} 15y_1 \equiv 1 \pmod{2} \implies y_1 = 1\\ 10y_2 \equiv 1 \pmod{3} \implies y_2 = 1\\ 6y_3 \equiv 1 \pmod{5} \implies y_3 = 1\\ \end{cases} $
Therefore, the four solutions are: $ \begin{cases} x_1 = 1.15.0 + 1.10.2 + 1.6.1 = 26\\ x_2 = 1.15.0 + 1.10.2 + 1.6.0 = 20\\ x_3 = 1.15.1 + 1.10.2 + 1.6.1 = 41\\ x_4 = 1.15.1 + 1.10.2 + 1.6.0 = 35\\ \end{cases} $
Edit
Add missing cases suggested by yunone
Case 5 $ \begin{cases} x \equiv 0 \pmod{2}\\ x \equiv 2 \pmod{3}\\ x \equiv 2 \pmod{5}\\ \end{cases} $ Case 6 $ \begin{cases} x \equiv 1 \pmod{2}\\ x \equiv 2 \pmod{3}\\ x \equiv 2 \pmod{5}\\ \end{cases} $
And the last two solutions are: $ \begin{cases} x_5 = 1.15.0 + 1.10.2 + 1.6.2 = 32 \equiv 2 \pmod{30}\\ x_6 = 1.15.1 + 1.10.2 + 1.6.2 = 47 \equiv 17 \pmod{30}\\ \end{cases} $
Am I in the right track? Any idea?
Thanks,