Can someone teach or explain to me Chinese Remainder Theorem?

This computer code incorporates the "Chinese Remainde Theorem + Modular multiplicative Inverse"
i=0;j=0;m=0;t=0;a=(4, 3, 5);r= (1, 1, 2);c=lcm(a); d=c / a[i];n=d % a[i] ;loop1:m++; if(n*m % a[i] ==1, goto loop, goto loop1);loop:s=(c/a[i]*r[j]*m);i++;j++;t=t+s;m=0;if(i< count a, goto4,m=m);printc,"m + ",t % c;return

ANSWER =60 m +  37, where m=0, 1, 2, 3......etc.

Look under this link for detailed explanation of CRT:
https://web2.0calc.com/questions/congruences

OMG THANK YOU SO MUCH! I just realized that the numbers are required to be relatively prime!
 

Thats why I got the problem wrong when I followed the video, because the Question had no relatively prime numbers! The question was making you guess and check.

I can solve system of modulos now! Thanks!

It is written in C++.

Thanks, probs during the summer (which is far far away lol) I am going to take some computing classes so I can be cool like you 

Probably I will start with Java, because it  is the most common and easy one.

THank you Guest and Nirvana!

Solution: 

This method uses Euler's totient and the modulo inverse functions.

\(\begin{array}{rcll} n &\equiv& {\color{red}1} \pmod {{\color{green}4}} \\ n &\equiv& {\color{red}1} \pmod {{\color{green}3}} \\ n &\equiv& {\color{red}2} \pmod {{\color{green}5}} \\ \text{Set } m &=& 4\cdot 3\cdot 5 = 60 \\ \end{array} \)

\(\small{ \begin{array}{l} n = {\color{red}1} \cdot {\color{green}3\cdot 5} \cdot \underbrace{ \underbrace{ \underbrace{ \underbrace{ \underbrace{ \underbrace{ [ { (\color{green}3 \cdot 5) }^{\varphi({\color{green}4}) -1 } \pmod {{\color{green}4}} ] }_{=\text{modulo inverse }(3\cdot 5) \mod 4 } }_{=(3\cdot 5)^{2-1} \mod {4}} }_{=(3\cdot 5)^{1} \mod {4}} }_{=(15\pmod{4})^{1} \mod {4}} }_{=(15)^{1} \mod {4}} }_{= 3} + {\color{red}1} \cdot {\color{green}4\cdot 5} \cdot \underbrace{ \underbrace{ \underbrace{ \underbrace{ \underbrace{ \underbrace{ [ { (\color{green}4\cdot 5) }^{\varphi({\color{green}3}) -1} \pmod {{\color{green}3}} ] }_{=\text{modulo inverse } (4\cdot 5) \mod {3}} }_{=(4\cdot 5)^{2-1} \mod {3}} }_{=(4\cdot 5)^{1} \mod {3}} }_{=(20\pmod{3})^{1} \mod {3}} }_{=(2)^{1} \mod {3}} }_{=2} + {\color{red}2} \cdot {\color{green}4\cdot 3} \cdot \underbrace{ \underbrace{ \underbrace{ \underbrace{ \underbrace{ \underbrace{ [ { (\color{green}4\cdot 3) }^{\varphi({\color{green} 5}) -1 } \pmod {{\color{green}5}} ] }_{=\text{modulo inverse } (4\cdot 3) \mod 5 } }_{=(4\cdot 3)^{4-1} \mod { 5}} }_{=(4\cdot 3)^{3} \mod {5}} }_{=(12\pmod{5})^{3} \mod {5}} }_{=(2)^{3} \mod {5}} }_{=3}\\ \\ n = {\color{red}1} \cdot {\color{green}3\cdot 5} \cdot [3] + {\color{red}1} \cdot {\color{green}4\cdot 5} \cdot [2] + {\color{red}2} \cdot {\color{green}4\cdot 3} \cdot [3] \\ n = 45+ 40 + 72 \\ n = 157 \\\\ n \pmod {m} = 157 \pmod {60} \\ n = 37+ k\cdot 60 \; \; k \in Z\\ \mathbf{n_{min}} \; \mathbf{=} \; \mathbf{37} \end{array} }\)

GA

What is the smallest whole number that has

a remainder of 1 when divided by 4,

a remainder of 1 when divided by 3,

and a remainder of 2 when divided by 5

So it becomes:

x = 1 mod(4)

x = 1 mod(3)

x = 2 mod(5)

\(\begin{array}{rcll} x &\equiv& {\color{red}1} \pmod {{\color{green}4}} \\ x &\equiv& {\color{red}1} \pmod {{\color{green}3}} \\ x &\equiv& {\color{red}2} \pmod {{\color{green}5}} \\ \text{Let } m &=& 4\cdot 3\cdot 5 = 60 \\ \end{array}\)

Because 4 and 3 and 5 are relatively prim  we can go on:

\(\begin{array}{|rclcl|} \hline x &=& {\color{red}1} \cdot {\color{green}3\cdot 5} \cdot \left[\dfrac{1}{\color{green}3\cdot 5}\pmod{4} \right] + {\color{red}1} \cdot {\color{green}4\cdot 5} \cdot \left[\dfrac{1}{\color{green}4\cdot 5}\pmod{3} \right] \\ &+& {\color{red}2} \cdot {\color{green}4\cdot 3} \cdot \left[\dfrac{1}{\color{green}4\cdot 3}\pmod{5} \right] + {\color{green}4\cdot 3\cdot 5}\cdot k \quad | \quad k \in \mathbb{Z} \\ \hline && \left[\dfrac{1}{\color{green}3\cdot 5}\pmod{4} \right] \\ && \equiv \left[ { (\color{green}3\cdot 5) }^{\varphi({\color{green}4}) -1 } \pmod {{\color{green}4}} \right] \quad | \quad \varphi({\color{green}4}) = 2 \\ && \equiv \left[ { 15 }^{2 -1} \pmod {{\color{green}4}} \right] \\ && \equiv \left[ { 15 } \pmod {{\color{green}4}} \right] \\ && \equiv \left[ { 3 } \pmod {{\color{green}4}} \right] \\ \hline && \left[\dfrac{1}{\color{green}4\cdot 5}\pmod{3} \right] \\ && \equiv \left[ { (\color{green}4\cdot 5) }^{\varphi({\color{green}3}) -1 } \pmod {{\color{green}3}} \right] \quad | \quad \varphi({\color{green}3}) = 2 \\ && \equiv \left[ { 20 }^{2 -1} \pmod {{\color{green}3}} \right] \\ && \equiv \left[ { 20 } \pmod {{\color{green}3}} \right] \\ && \equiv \left[ { 2 } \pmod {{\color{green}3}} \right] \\ \hline && \left[\dfrac{1}{\color{green}4\cdot 3}\pmod{5} \right] \\ && \equiv \left[ { (\color{green}4\cdot 3) }^{\varphi({\color{green}5}) -1 } \pmod {{\color{green}5}} \right] \quad | \quad \varphi({\color{green}5}) = 4 \\ && \equiv \left[ { 12 }^{4 -1} \pmod {{\color{green}5}} \right] \\ && \equiv \left[ { 12^3 } \pmod {{\color{green}5}} \right] \quad 12 \pmod{5} \equiv 2 \pmod{5} \\ && \equiv \left[ { 2^3 } \pmod {{\color{green}5}} \right] \\ && \equiv \left[ { 8 } \pmod {{\color{green}5}} \right] \\ && \equiv \left[ { 3 } \pmod {{\color{green}5}} \right] \\ \hline x &=& {\color{red}1} \cdot {\color{green}3\cdot 5} \cdot [3] + {\color{red}1} \cdot {\color{green}4\cdot 5} \cdot [2] + {\color{red}2} \cdot {\color{green}4\cdot 3} \cdot [3] + 60\cdot k \quad | \quad k \in \mathbb{Z} \\ x &=& 45+ 40 + 72 + 60\cdot k \\ x &=& 157+ 60\cdot k \quad | \quad 157 \pmod {60} = 37 \\ x &=& 37+ 60\cdot k \qquad k \in Z \\ \mathbf{x_{min}} &=& \mathbf{ 37} \\ \hline \end{array}\)

wow... this site has shown its true power!

Thank you guys!