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- How to Prove the divisibility rule for $3$ 17 answers
When you add the digits of any number that is divisible by three, that sum of those digits also appears to be divisible by three (with no remainder).
For example a number (which I randomly grab from the top of my head):
whose digits sum to 33 (2+8+9+7+5+2=33)
That sum 33 is divisible by three and so is the original number 289752.
This is not the case when dividing by 2, for example 12 is divisible by two but when its digits are summed (1+2=3) you receive 3 which is NOT divisible by 2.
I have yet to be able to find a counter example for this phenomenon of division by 3.
Why does this happen?