Working of divisibility for 3 and 9

We are aware that when any number has its unit digit as 0, 2, 4, 6, or 8 (all even numbers), the number is said to be divisible by 2. Similarly, to check if a number is divisible by 3 or 9, we simply sum up the digits of the number and see if it’s divisible by 3 or 9. If yes, then the number is divisible by 3 or 9 respectively.

For example, let’s take a random number N = 45831, and see if it’s divisible by 3 or not. 

Sum up the digits of the number 45831
=> 4 + 5 + 8 + 3 + 1 = 21
21 is divisible by 3? YES (3 x 7 = 21)
Hence, 45831 is divisible by 3
21 is divisible by 9? NO
Thus, 45831 is not divisible by 9

Divisibility is when you can divide the number evenly by a factor, leaving no remainder. If we take a real life example, a pizza having 4 slices, can be evenly distributed between 2 people, while it cannot be evenly given between 3 people. Here we say, 4 is divisible by 2. Divisibility rules help you work with large numbers to find their factors, and determine if the numbers are prime or composite.

Ever wondered how and why the divisibility formula works?

Well, I didn’t until I read about it recently. Let me share with you the working behind the strategy. As you read further, you’ll see that the propositions used, are all based on number theories, arithmetic laws & properties. 

We start with basic representation of numbers. Consider a number 4708492, the 3 common ways of expressing the number are: 

  • Standard form: 4,708,492 
  • Expanded form: 4 x (1,000,000) + 7 x (100,000) + 0 x (10,000) + 8 x (1000) + 4 x (100) + 9 x (10) + 2 x (1) 
  • Words form: Four million, seven hundred eight thousand, four hundred, and ninety two 
N = 4708492, here, we use the expanded form, and rewrite it as:  
4 x (106) + 7 x (105) + 0 x (104) + 8 x (103) + 4 x (102) + 9 x (101) + 2 x (100
Similarly, we can take any number N = abcde, and write it in its expanded form as:
a x (104) + b x (103) + c x (102) + d x (101) + e x (100
which is same as: 
104a + 103b + 102c + 101d + 100e  =>  104a + 103b + 102c + 10d + e 
We write 10=9+1, and we get… 
(9+1)4a + (9+1)3b + (9+1)2c + (9+1)d + e  
Further, using the distributive laws and assuming all the multiples of 9 as m, we write 
[9m + (1)4]a + [9m + (1)3]b + [9m + (1)2]c + [9m + (1)]d + e 
Factoring out the common term 9m, we get
N = 9m [a + b + c + d] + a + b + c + d + e 

Through the above equation, theories conclude that for N to be divisible by 3, if (a + b + c + d + e) is divisible by 3 then, N is divisible by 3. Same logic applies for divisibility by 9!

Let’s take some examples for both 3 and 9 and see how the equation unfolds…

N= 45831 
N = 4 x (104) + 5 x (103) + 8 x (102) + 3 x (101) + 1 x (100
= 4 x (9+1)4 + 5 x (9+1)3 + 8 x (9+1)2 + 3 x (9+1) + 1 
= 4 [9m + (1)4] + 5 [9m + (1)3] + 8 [9m + (1)2] + 3 [9m + (1)]d + 1 
= 9m [4 + 5 + 8 + 3] + 4 + 5 + 8 + 3 + 1 
= 9m [20] + 21 
= 9m + 3m
= 3m(3m+1) 
=  3[m] 
N = 45831 is a multiple of 3, hence is divisible by 3 but not by 9 
For N = 45837 
N = 4 x (104) + 5 x (103) + 8 x (102) + 3 x (101) + 7 x (100
= 4 x (9+1)4 + 5 x (9+1)3 + 8 x (9+1)2 + 3 x (9+1) + 7 
= 4 [9m + (1)4] + 5 [9m + (1)3] + 8 [9m + (1)2] + 3 [9m + (1)]d + 7 
= 9m [4 + 5 + 8 + 3] + 4 + 5 + 8 + 3 + 7 
= 9m [20] + 27 
= 9m + 9m
= 9(2m) 
= 9[m]  
Here, N = 45837 is a multiple of 9, and also of 3, hence is divisible by 9 as well as 3 
For N = 63
= 6 x 10 + 3 x 1
= 6 [9+1] + 3
= 6 [9] + 6 + 3
= 9(6) + 9
= 9 (6+1)
= 9(7)
=> 63 is divisible by 9 and 3

We saw the working behind the divisibility for 3 and 9, similarly, there exists respective reasoning for other divisibility checks as well. The more we dig into the nuances of math concepts, the more magical it appears.

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