<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 $:$

  
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*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] Since we are working in a numerical system of basis $10$, an integer number $N>0$ can be written as $$N=a_n\times 10^n+a_{n-1}\times 10^{n-1}+\ldots +a_1\times 10 +a_0$$
The number $10=9+1$ leaves remainder $1$ under division by $3$, and so does the another powers of $10$, so $N$ has the form $$N=9m+a_{n}+a_{n-1}+\ldots+a_1+a_0$$
for some integer $m$, which means that $3$ divides $N$ iff $3$ divides $a_{n}+a_{n-1}+\ldots+a_1+a_0$, the sum of the digits of $N$. Also the same test works for $9$.
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