Why does the summation of the digits of a number divisible by 3 also become divisible by 3?
In: Mathematics
Here’s an algebraic explanation, to add to the others.
Let’s use T to represent the tens place of a two digit number, and O to represent the ones place.
So any two digit number can be written as:
10T + O
for example, 42 = 10 x 4 + 2
With me so far?
10 = 9 + 1 so we can write
10T + O = (9 + 1)T + O = 9T + T + O
Now we say this number is the result of multiplying 3 by another number, which we’ll call N
3N = 9T + T + O
We know 9T is divisible by 3 because 9 is divisible by 3. In order for the whole number to be divisible by 3, then T + O must also be divisible by 3. And T + O is just the sum of the digits!
This is for a two digit number, but you can extend it to any number of digits, because you can expand any number into powers of 10, which can then be expanded to (9 + 1)^n which expands to a bunch of numbers that are multiples of 9, plus a 1 at the end. Getting rid of the 9 terms, which we know are divisible by 3, it leaves us with a bunch of 1s which multiply the digits, resulting in the sum of the digits.
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