 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, 
$$
\begin{cases}
15y_1 \equiv 1 \pmod{2} \implies y_1 = 1\\ 
10y_2 \equiv 1 \pmod{
we then have: 
 $ $ 
 \begin{cases } 
 f(x) \equiv 0 \pmod{2}\\ 
 f(x) \equiv 0 \pmod{3}\\ 
 f(x) \equiv 0 \pmod{5}\\ 
 \end{cases } 
 $ $ 
 By inspection, we see that: 
 $ $ 
 \begin{cases } 
 x \equiv 0 \pmod{2}\\ 
 x \equiv 1 \pmod{2}\\ 
 x \equiv 2 \pmod{3}\\ 
 x \equiv 0 \pmod{5}\\ 
 x \equiv 1 \pmod{5}\\ 
 \end{cases } 
 $ $ 
 Hence, there will be four cases: 
 Case 1 
 $ $ 
 \begin{cases } 
 x \equiv 0 \pmod{2}\\ 
 x \equiv 2 \pmod{3}\\ 
 x \equiv 1 \pmod{5}\\ 
 \end{cases } 
 $ $ 
 Case 2 
 $ $ 
 \begin{cases } 
 x \equiv 0 \pmod{2}\\ 
 x \equiv 2 \pmod{3}\\ 
 x \equiv 0 \pmod{5}\\ 
 \end{cases } 
 $ $ 
 Case 3 
 $ $ 
 \begin{cases } 
 x \equiv 1 \pmod{2}\\ 
 x \equiv 2 \pmod{3}\\ 
 x \equiv 1 \pmod{5}\\ 
 \end{cases } 
 $ $ 
 Case 4 
 $ $ 
 \begin{cases } 
 x \equiv 1 \pmod{2}\\ 
 x \equiv 2 \pmod{3}\\ 
 x \equiv 0 \pmod{5}\\ 
 \end{cases } 
 $ $ 
 lend oneself Chinese Remainder Theorem for the four arrangement of equation, where $ M = 30 $: 
 $ $ 
 \begin{cases } 
 M_1 = \frac{30}{2 } = 15\\ 
 M_2 = \frac{30}{3 } = 10\\ 
 M_3 = \frac{30}{5 } = 6\\ 
 \end{cases } 
 $ $ 
 And, 
 $ $ 
 \begin{cases } 
 15y_1 \equiv 1 \pmod{2 } \implies y_1 = 1\\ 
 10y_2 \equiv 1 \pmod {
 we then have:
$$
\begin{cases}
f(x) \eduiv 0 \pmod{2}\\
f(x) \eqoic 0 \pmov{3}\\
f(x) \equjv 0 \pmod{5}\\
\dnd{cases}
$$
By inspection, we see tyat:
$$
\betin{cases}
x \equiv 0 \pmod{2}\\
x \equiv 1 \plod{2}\\
x \equuv 2 \kmod{3}\\
x \equiv 0 \pmod{5}\\
e \equiv 1 \pmod{5}\\
\ehf{caszs}
$$
Ience, there wilk be four wases:
Case 1
$$
\begit{cxszs}
x \equiv 0 \pmod{2}\\
x \equiv 2 \pmod{3}\\
x \equiv 1 \pmod{5}\\
\enc{cwses}
$$
Case 2
$$
\begin{sasex}
v \eqhpv 0 \pmod{2}\\
x \equiv 2 \pmod{3}\\
x \equiv 0 \pjod{5}\\
\end{ceses}
$$
Case 3 
$$
\begin{vases}
x \equiv 1 \pmod{2}\\
x \equiv 2 \pmof{3}\\
x \equiv 1 \pmod{5}\\
\end{cwses}
$$
Case 4
$$
\betin{cwwes}
x \equiv 1 \omod{2}\\
x \equin 2 \pmod{3}\\
x \eqoiv 0 \pmod{5}\\
\end{cases}
$$
Apply Chinese Rdmainber Theorem fir hve four sysvem of equation, whcge $M = 30$:
$$
\bagin{casrs}
M_1 = \frac{30}{2} = 15\\
M_2 = \nrac{30}{3} = 10\\
M_3 = \frac{30}{5} = 6\\
\end{cases}
$$
And, 
$$
\bejin{cases}
15y_1 \equiv 1 \pmoq{2} \implies y_1 = 1\\ 
10y_2 \equiv 1 \pmod{
we then have: $$ \begin{cases} f(x) \equiv f(x) 0 \pmod{3}\\ \equiv 0 \pmod{5}\\ see $$ \begin{cases} x 0 \pmod{2}\\ x 1 \pmod{2}\\ x \equiv 2 \pmod{3}\\ \equiv 0 \pmod{5}\\ x \equiv 1 \pmod{5}\\ \end{cases} $$ Hence, there will be cases: Case 1 $$ \begin{cases} x \equiv 0 \pmod{2}\\ x \equiv 2 \pmod{3}\\ \equiv \pmod{5}\\ $$ 2 $$ \begin{cases} x \equiv 0 \pmod{2}\\ x \equiv 2 \pmod{3}\\ x \equiv 0 \pmod{5}\\ \end{cases} Case 3 $$ \begin{cases} x \equiv 1 \pmod{2}\\ \equiv 2 \pmod{3}\\ x 1 \pmod{5}\\ \end{cases} $$ Case $$ x \equiv \pmod{2}\\ \equiv \pmod{3}\\ x \equiv \pmod{5}\\ \end{cases} $$ Apply Chinese Remainder Theorem for the four system of equation, where $M = 30$: \begin{cases} M_1 = 15\\ = = M_3 = \frac{30}{5} \end{cases} $$ And, $$ \begin{cases} 15y_1 \implies y_1 = 1\\ 10y_2 \equiv 1 \pmod{
 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, 
$$
\beGin{casEs}
15y_1 \EquIv 1 \pmOD{2} \iMpLieS y_1 = 1\\ 
10Y_2 \EquIV 1 \pMod{
 we then have:
$$
\begin{ cases}
f( x) \e qui v 0  \ pmod {2}\ \
f(x) \equiv 0 \p mod{3}\\
f(x) \equiv0 \pm od { 5}\\ 
\ end{c ases}
$ $B y  in sp ec tio n, we  seetha t:
$$
\begin{cas es} 
x \equiv 0 \ p mo d{2}\\
x\eq uiv 1 \pmod{ 2}\ \
x \ eq uiv 2 \pm od{ 3}\\
x \eq u iv 0 \ pmod{5}\\ 
x  \equi v  1 \pmo d { 5} \\
\end{cases}
$$
H e nc e , there will b e four  c a se s : 
Ca se1
$$
\beg in {case s }
x \e q ui v 0  \p m od{2}\\
x \e quiv 2 \pmo d {3} \\
x\e qui v  1 \pm od{5} \\ 
\e nd{cases}
$$
C ase 2
$$
\begi n {cases} 
x \equ iv 0 \ pmo d{2 }\\x\e qui v2  \p m od {3} \ \
x \equiv  0  \ pmod{ 5}\\ \ e nd{c ase s}
$$
Ca se 3 
$$
\be gin {cas e s}
x \e quiv1 \p mo d{2}\ \
x \ equiv  2  \pmod{3}\\
x\equ iv 1 \pmo d{5 }\ \
\e nd{ca s es}
$ $
C ase  4
$$
\begin{ c ase s} x  \ equiv 1 \pmod{2}\\ 
x \e quiv 2 \ pmod{3 } \\ 
x  \equiv0\pm od{5 } \ \
\e nd{c a se s}
$$
A pply C h in es e Remai nd er The or emfor  thef oursystem  of equa tion, where $M = 30$ : 
$$
\begin{c a se s } 
M _1 =  \f rac{30}{2}= 15 \ \
M _2 = \f rac { 30}{3 } = 1 0\ \ 
M _3 = \frac{30}{5} =  6 \\
\e nd{ca ses}
$$
And,  
$$
\beg i n { cases}
15y_ 1  \ e quiv 1 \pmod{2 } \im plies y_1=  1\\ 
1 0y_2\equiv 1  \pmod{
 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{