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??

• Do you know any properties of numbers that are divisible by 3? – Mc Cheng Jun 6 '16 at 11:56
• They are also divisible by 9. – ncmathsadist Jun 6 '16 at 11:58
• hmmm. no I will try to research some. – Can't Read Fast Jun 6 '16 at 12:00
• A number is divisible by 3 if and only if the sum of its digits is divisible by 3. The kind of people who answer questions on this site find it so hard to believe that you have not already met this that they are reluctant to give you a straightforward answer :) – almagest Jun 6 '16 at 12:03
• i guess i learned some thing today :D – Can't Read Fast Jun 6 '16 at 12:09

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.

• 10 factorial is not the number you mean. – Bennett Gardiner Jun 10 '16 at 7:20
• 10! numbers can be formed using numbers 0 to 9 without repeating the nos. – Amruth A Jun 10 '16 at 7:26

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 $3$.

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.