- #1
Char. Limit
Gold Member
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Why is it that if the sum of the digits of a number is divisible by three, the number itself is also divisible by three?
I've tried to do it, but I can't get anywhere. This is the best I have.
Start with an integer x, which can also be written thus...
[tex]x=x_1+x_2\times10+x_3\times10^2+x_4\times10^3+...+x_i\times10^{i-1}[/tex]
Which gives us the digits of x. So, why is it that if...
[tex]\frac{\sum_{i=1}^{n} x_i}{3} \in \mathbb{Z}[/tex]
then...
[tex]\frac{x}{3} \in \mathbb{Z}[/tex]
for an n-digit number x?
I've tried to do it, but I can't get anywhere. This is the best I have.
Start with an integer x, which can also be written thus...
[tex]x=x_1+x_2\times10+x_3\times10^2+x_4\times10^3+...+x_i\times10^{i-1}[/tex]
Which gives us the digits of x. So, why is it that if...
[tex]\frac{\sum_{i=1}^{n} x_i}{3} \in \mathbb{Z}[/tex]
then...
[tex]\frac{x}{3} \in \mathbb{Z}[/tex]
for an n-digit number x?