Divisibility properties are best expressed using modular arithmetic. I’ll save on the notation but the idea is as follows:

Consider a number like `ABCD` where the letters are the digits. The number itself is equal to 1000 A + 100 B + 10 C + D. Whereas the sum of digits is A + B + C + D. We claim that these have the same remainder when divided by 3.

Well this is not hard to see because the difference – 999 A + 99 B + 9 C – is divisible by 3 and so has remainder 0 when divided by 3. Thus `ABCD` is divisible by 3 if and only if its sum of digits is.

This argument can easily be extended to all numbers (it’s just cumbersome to write on reddit). And by replacing 3 with 9 we get the same rule for 9.

It’s just a natural pattern, which you can see counting upwards all multiples of 3.

* 03 – 06 – 09 … all divisible by 3

* 12 – 15 – 18 … notice that, you have +1 in the tens place, and -1 in the one’s place

* 21 – 24 – 27 … +2 in the tens plance, -2 in the one’s place

* 30 – 33 – 36 – 39 … and now the pattern repeats

This works in the base 10 system because it cycles cleanly over the ten’s place, always maintaining this relationship between the rightmost digits.

Now to further expand this, and you’ll note the hundreds and tens place follow the same (extended) pattern.

* 000 – 003 – 006 – 009

* …

* 030 – 033 – 036 – 039

* …

* 060 – 063 – 066 – 069

* …

* 090 – 093 – 096 – 099

* 102 – 105 – 108

* 111 – 114 – 117 … +1 hundred’s place, +1 ten’s place, -2 one’s place

* 120 – 123 – 126 – 129 … +1 hundred’s place, +2 ten’s place, -3 (or 0) one’s place

Thus, every increase in a value to the digit to the left, is accompanied by an decrease in the value of the one’s place. In this way, they balance out, so whatever’s left if you add them to the one’s place will *always* be a multiple of 3.

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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