Ten digit numbers divisible by 3. I came across an interesting property of 10-digit numbers that are constructed using each digit only once: e.g. $9867534210$ or $352147890$. These numbers are exactly divisible by $3$. Each and every of the $10!$ combinations are also divisible by $3$.
But why is this property emerging, i have no idea. Can somebody explain this to me why this happens??
 A: Because a number is divisible by 3 if the sum of the digits in number is divisible by 3.
Since sum of 0+1+2+3+4+5+6+7+8+9=45 and 45 is divisible by 3.
All the possible numbers formed of 10! will be divisible by 3. 
A: You have $10$ digits $[0,1,2,3,4,5,6,7,8,9]$ and if you construct any possible number by taking each digits once,you'll get $10!$ numbers.

A number is divided by $3$,if sum of the digits of the number is divided by $3$

For all these numbers (Sum of digits)=$(0+1++2+3+4+5+6+7+8+9)=45$ is divided by $3$.
So,all these number are divisible by $3$.
A: To know Mod(x,9) or Mod(x,3), we can substitute x with sum of digits of $x$ in decimal
Proof:
For any integer $x$, $10x - x = 9x$
since $9x$ is times of 9: $Mod(x,9) = Mod(10x,9)$
Further: $Mod(x,9)=Mod(10x,9)=Mod(100x,9) ...$
Using the properties above, for any integer expressed in decimal form:
$x=x_0+x_1*10+x_2*100+x_3*1000 ...$
Since $Mod(a+b,c)=Mod(Mod(a,c)+Mod(b,c),c)$
We have:
$Mod(x,9)$
$=Mod(Mod(x_0,9)+Mod(x_1*10,9)+Mod(x_2*100,9)...,9)$
$=Mod(Mod(x_0,9)+Mod(x_1,9)+Mod(x_2,9)...,9)$
$=Mod(x_0+x_1+x_2...,9)$
With 9 case proven, 3 case is immediately proven.
