<s>[INST] Why test of divisible by $12$ works with $3$ and $4$ but not with $2$ and $6 ?$ Test of divisible by $4 ,$ last two digit must be divisible by $4 ,$ since $100$ is always divisible by $4$ remaining two digit $,$ we need to check $.$
Test of divisible by $3 ,$ sum digits must be divisible by $3 .$ But I don't know why $?$
Test of divisible by $2 ,$ last digit must be divisible by $2 ,$ since $10$ is always divisible by $2 .$
Test of divisible by $ 6 ,$ the number should be divisible by $2$ and $3 .$
Test of divisible by $12 ,$ the number should be divisible by $4$ and $3 ,$ somewhere  is given as we can say if number is divisible by $2$ and $6$ then it also must be divisible by $12 ,$ for counter example $,$ if number is $18 .$

I have two questions $:$

  
*
  
*Why test of divisible by $3$ works $?$ Any proof $.$
  
*Why test of divisible by $12$ works with divisible by $3$ and $4 ,$ but not with divisible by $2$ and $6 ?$ Any proof $.$
  

 [/INST] The reason the divisibility by $3$ test works is that $10$ and $1$ have the same remainder when divided by $3$. Say your number is (as digits) $x = a_n a_{n-1} \ldots a_1 a_0$. Then you could write it as $x = a_n \cdot 10^n + a_{n-1} 10^{n-1} + \cdots + a_1 10 + a_0$. If you mod out by $3$:
$$
\begin{align*}
x \bmod 3 &= (a_n \cdot 10^n + a_{n-1} 10^{n-1} + \cdots + a_1 10 + a_0) \bmod 3 \\
&= (a_n \cdot 10^n) \bmod 3 + (a_{n-1} 10^{n-1}) \bmod 3 + \cdots + (a_1 10) \bmod 3 + a_0 \bmod 3) \\
&= (a_n \bmod 3) \cdot (10^n \bmod 3) + (a_{n-1} \bmod 3) (10^{n-1} \bmod 3) + \cdots + (a_1 \bmod 3) \cdot (10 \bmod 3) + (a_0 \bmod 3) \\
&= (a_n \bmod 3) \cdot 1 + (a_{n-1} \bmod 3) \cdot 1 + \cdots + (a_1 \bmod 3) \cdot 1 + (a_0 \bmod 3) \\
&= (a_n \bmod 3) + (a_{n-1} \bmod 3) + \cdots + (a_1 \bmod 3) + (a_0 \bmod 3) \\
&= (a_n + a_{n-1} + \cdots + a_1 + a_0) \bmod 3
\end{align*}
$$
So $x$ and the sum of its digits have the same remainder when divided by $3$.
As for your second question: if you know $x$ is divisible by $6$, you already know it's divisible by $2$. Doing the test for $2$ doesn't give you any new information. This is because $6$ and $2$ aren't relatively prime (but $3$ and $4$ are). In general, if you know $x$ is divisible by $a$ and $b$, it's divisible by the LCM of $a$ and $b$.
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