How to prove that the sum of each digit of a number can be divided by 3, then the number can also be divided by 3? Please show me the process, don't give me an example But what I want is a general formula, not an example. If it's five digits or six digits (I say a number, you can't be sure it's four digits), you can't prove it in this way

A=a0+10a1+10^2a2+10^3a3+…… =[(10-1)a1+(10^2-1)a2+(10^3-1)a3+…… ]+(a0+a1+a2+a3+……） It is easy to check that 10 ^ n-1 (n is a natural number) is a multiple of 3 and 9. So we can draw a conclusion that whether a is a multiple of 3 or 9, just look at the number of a and A0 + A1 + A2 + a3 + Is it 3 or 9

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