Tests of Divisibility

This topic gives an overview of;

- Tests of Divisibility
- Divisibility by 10
- Divisibility by 5
- Divisibility by 2
- Divisibility by 9 and 3

In previous classes, you learnt how to check divisibility by the following divisors.

10, 5, 2, 3, 6, 4, 8, 9, 11.

You would have found the tests easy to do, but you may have wondered at the same time why they work.

This is certainly the easiest test of all! We first look at some multiples of 10 such as 10, 20, 30, 40, 50, 60, ... , and then at some non-multiples of such as 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.

Look at the multiples of 5 such as 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 + a*

Since 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.

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

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

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

Look carefully at the three tests of divisibility found till now, for checking division by 10, 5 and 2. We see something common to them: *they use only the one’s digit of the given number; they do not bother about the ‘rest’ of the digits*. Thus, divisibility is decided just by the one’s digit. 10, 5, 2 are divisors of 10, which is the key number in our place value.

But for checking divisibility by 9, this will not work. Let us take some number say 3573.

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

This is equal to3 × (999 + 1) + 5 × (99 + 1) + 7 × (9 + 1) + 3

= 3 × 999 + 5 × 99 + 7 × 9 + (3 + 5 + 7 + 3) -----(1)

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. As above, we get 3576 = 3 × 999 + 5 × 99 + 7 × 9 + (3 + 5 + 7 + 6) ... (2)

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,

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

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

- A number is said to be divisible by another number, when the remainder is zero.
- A number is divisible by 10, if its ones digit is 0.
- A number is divisible by 5, if its ones digit is 0 or 5.
- A number is divisible by 2, if its ones digit is 0, 2, 4, 6 or 8.
- If a number is divisible by 10, then the number is also divisible by 2 and 5.
- A number is divisible by 9, if the sum of its digits is divisible by 9.
- A number is divisible by 3, if the sum of its digits is divisible by 3.
- If a number is divisible by 9, then the number is also divisible by 3.
- If a number is divisible by 3, then it may not necessarily be divisible by 9.

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