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:
- the sum of any number to 9 preserve the sum of its digits
- 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.