In this blog, we will discuss various divisibility test rules.

Divisibility Test of 10

This is certainly the easiest test of all! We first look at some multiples of 10. 

10, 20, 30, 40, 50, 60, ... and so on, and at some non-multiples of 10.

13, 27, 32, 48, 55, 69

From these lists we see that if the ones digit of a number is 0, then the number is a multiple of 10; and if the ones digit is not 0, then the number is not a multiple of 10. So, we get a test of divisibility by 10.

Of course, we must not stop with just stating the test; we must also explain why it �works�. That is not hard to do; we only need to remember the rules of place value.

Take the number. ... cba; this is a short form for ... + 100c + 10b + a. Here a is the one�s digit, b is the ten�s digit, c is the hundred�s digit, and so on. The dots are there to say that there may be more digits to the left of c. Since 10, 100, ... are divisible by 10, so are 10b, 100c, ... . And as for the number a is concerned, it must be a divisible by 10 if the given number is divisible by 10. This is possible only when a = 0.

Hence, a number is divisible by 10 when its one�s digit is 0.

Divisibility Test of 5

Look at the multiples of 5.

5, 10, 15, 20, 25, 30, 35, 40, 45, 50,

We see that the one�s digits are alternately 5 and 0, and no other digit ever appears in this list.

So, we get our test of divisibility by 5.

If the ones digit of a number is 0 or 5, then it is divisible by 5.

Let us explain this rule. Any number ... cba can be written as:

... + 100c + 10b + aSince 10, 100 are divisible by 10 so are 10b, 100c,� ... which in turn, are divisible by 5 because 10 = 2 � 5. As far as number �a� is concerned it must be divisible by 5, if the number is divisible by 5. So �a� has to be either 0 or 5.

Divisibility Test of 2

Here are the even numbers.

2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, ...  and here are the odd numbers.

1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, ... ,

We see that a natural number is even if its one�s digit is

2, 4, 6, 8 or 0

A number is odd if its one�s digit is

1, 3, 5, 7 or 9

The test of divisibility by 2 is as follows.

If the one�s digit of a number is 0, 2, 4, 6 or 8 then the number is divisible by 2.

The explanation for this is as follows.

Any number cba can be written as 100c + 10b + a

The first two terms namely 100c, 10b are divisible by 2 because 100 and 10 are divisible by 2. So far as �a� is concerned, it must be divisible by 2 if the given number is divisible by 2. This is possible only when a = 0, 2, 4, 6 or 8.

Divisibility Tes of 9 and 3

Let us take some number say 3573.

Its expanded form is: 3 � 1000 + 5 � 100 + 7 � 10 + 3

This is equal to 3 � (999 + 1) + 5 � (99 + 1) + 7 � (9 + 1) + 3 = 3 � 999 + 5 � 99 + 7 � 9 + (3 + 5 + 7 + 3)

We see that the number 3573 will be divisible by 9 or 3 if (3 + 5 + 7 + 3) is divisible by 9 or 3.

We see that 3 + 5 + 7 + 3 = 18 is divisible by 9 and also by 3. Therefore, the number

3573 is divisible by both 9 and 3.

Now, let us consider the number 3576. Its expanded form is: 3 � 1000 + 5 � 100 + 7 � 10 + 3 This is equal to 3 � (999 + 1) + 5 � (99 + 1) + 7 � (9 + 1) +3

3576 = 3 � 999 + 5 � 99 + 7 � 9 + (3 + 5 + 7 + 6) 

Since (3 + 5 + 7 + 6) i.e., 21 is not divisible by 9 but is divisible by 3,

therefore 3576 is not divisible by 9. However, 3576 is divisible by 3.

Hence,

(i) A number N is divisible by 9 if the sum of its digits is divisible by 9. Otherwise it is not divisible by 9.

(ii) A number N is divisible by 3 if the sum of its digits is divisible by 3. Otherwise it is not divisible by 3.

If the number is �cba�, then, 100c + 10b + a = 99c + 9b + (a + b + c)

= 9(11c + b) +(a + b + c) where 9(11c + b) divisible by 3 and 9

Hence, divisibility by 9 (or 3) is possible if a + b + c is divisible by 9 (or 3).