The divisibility rule of 3

While reviewing my daughter's homework, I got to think a little bit about the divisibility rule of 3. Actually, the basic operations have some interesting properties:

1. the sum of any number to 9 preserve the sum of its digits
2. any number multiplied by 3 results in a number which the sum of its digits will also be a multiple of 3

In this post, I'm going to analyze a little bit the reasons behind each of these properties.

1) the sum of any number to 9 preserve the sum of its digits

Considering the operation below

\[A + 9 = B\]

The sum of digits of A will be equal to the sum of digits of B

Take a look at some examples:

Example 1.1:

\[15 + 9 = 24\]

the sum of the digits of \(15\) will be equal to the sum of the digits of the resulting value \(24\) (\(2 + 4 = 6\))

Example 1.2:

\[71 + 9 = 80\]

\(7 + 1\) is equal to \(8 + 0\), both \(8\)

2) any number multiplied by 3 results in a number which the sum of its digits will also be a multiple of 3

Considering the operation

\[A * 3 = B\]

this property implies that the sum of the digits of B will necessary be a multiple of \(3\). As we can see in the examples below:

Example 2.1:

\[24 * 3 = 72\]

As we can notice the sum of the digits of \(72\) is \(7 + 2 = 9\), a multiple of 3

Example 2.2:

\[19*3 = 57\]

the sum of digits of \(57\) is \(5 + 7 = 12\), also a multiple of 3

A deep dive of the divisibility rule of 3

In elementary school we learn the divisibility rule of 3 and accept that it happens without really understanding why.

TIL that the property 2 is derived of property 1, i.e, when multiplying a number by \(3\) it also means a sum with \(9\).

Wait, what?

Well, the multiplication is actually a sequence of sum operations, right?

So a simple operation like

\[3 * 4 = 12\]

can also be expressed as a sequence of sum

\[3 + 3 + 3 + 3 = 12\]

and can be rearranged to be a sum with \(9\)

\[9 + 3 = 12\]

Let me show other examples in which multiplication by 3 reveals to be a sum with \(9\)

Example 3.1

\[3*5 = 3+3+3+3+3 = 9+6 = 15.\]

Example 3.2

\[3*6 = 3+3+3+3+3+3 = 9+9 = 18\]

Multiplying by 3 means a sum between a number k plus 9

Hence, we can conclude that the property 1 is the reason behind property 2.

As the multiplication by \(3\) implies a sum between a number \(k\) plus \(9\), the property 1 garantees that the sum of the digits of \(k\) will be preserved and as \(k\) is a multiple of 3, the sum of the digits of the result will also be a multiple of 3.

Totally Crazy, right?

Below we can see a proof:

Considering any real number \(x > 3\) and a \(k = x - 3\):

\[3*x = 3*(k+3) = 3*k+9\]

As \(3 * k\) is a multiple of 3, by the rules of property 1, it's implied that the digits of \(3 * k + 9\) will also be multiple of \(3\).

What's the reason behind property 1?

It's worth to note that it happens because to sum with \(9\) is the same thing to add \(10\) and to subtract \(1\) at same time, i.e., increasing one in second digit and decreasing one in first digit, what does not change the sum of digits ($ +1 - 1 = 0$).

We can also generalize that the property 2 happens with all numeral system with the following number of digits:

\[3 * N + 1\]

for instance, decimal and hexadecimal numeral systems.

And the number that follows the property 1 is always the maximum digit of the numeral system minus \(1\): so it is \(9\) for decimal numeral system and \(F\) for hexadecimal numbers.