There are a number of divisibility rules to determine whether one number is divisible by another number.

Often, we have to divide two numbers, and while we can obtain the answers through ‘first principles’ i.e. by solving the problem step-by-step, it can be cumbersome and time consuming.

A number is said to be divisible by:

- 2 if it an even number i.e. it ends with 0, 2, 4, 6 or 8.
- 3 if the sum of its digits is divisible by 3
- 4 if the last 2 digits of the number are divisible by 4
- 5 if the last digit (i.e. the units digit) ends with 5 or 0
- 6 if the number is divisible by both 2 and 3
- 8 if the number formed by the last three digits is divisible by 8
- 9 if the sum of the digits is divisible by 9
- 10 if the last digit (i.e. units digit) is 0
- 11 if the sum of the digits in the odd position is equal to the sum of the digits in the even position OR they differ by a number that is divisible by 11

You would have noticed that 7 is not in the above list, and it is because the rules are not always practical to apply. But here it is anyway.

To check for a number’s divisibility by 7, we leave out the units digit, and obtain a truncated number (the number formed by all digits *other than* the unit digit). We multiply the truncated number by 3, and then add the units digit to the product. If the sum of these two numbers is divisible by 7, then the original number is divisible by 7.

Let’s look at an example to test the divisibility of the number 126 by 7.

- Leave the units digit of ‘6’. The truncated number is 12.
- Truncated number multiplied by 3 i.e. 12 x 3 = 36
- Add the units digit to the product obtained above i.e. 36 + 6 = 42
- 42 is divisible by 7.
- Hence 126 is divisible by 7 (126 ÷ 7 = 18)

Now let us look at another example to test the divisibility rules for the number 297.

- 297 is not an even number, so it is not divisible by 2
- 2 + 9 + 7 = 18. 18 ÷ 3 = 6. So 297 is divisible by 3
- Last 2 digits of the number are 97, and it is not divisible by 4 (297 not divisible by 4)
- Last digit is 7, not divisible by 5
- 297 is not even, not divisible by 2. But it is divisible by 3. So 297 is not divisible by 6 since it is not divisible by
*both*2 and 3. - 297 ÷ 8 = 37 with a remainder of 1, so not divisible by 8
- 2 + 9 + 7 = 18. 18 ÷ 9 = 2. So 297 is divisible by 9
- Not ending in 0, hence not divisible by 10
- 2 + 7 = 9 (sum of odd digits) = to the sum of even digits (9 here). 279 is divisible by 11
- Truncated number = 29. (29 x 3) + 7 = 87 + 7 = 94. 94 ÷ 7 = 13r2. Hence 279 is not divisible by 7.

We now look at the divisibility rules for 14 and 15.

A number is divisible by 14 if it is divisible by *both* 2 and 7.

Let’s take 126. It is even, so it is divisible by 2. Is it divisible by 7?

(12 x 3) + 6 = 36 + 6 = 42 ÷ 7 = 6 (divisible by 7).

So 126 is divisible by 14

A number is divisible by 15 if it is divisible by *both* 3 and 5.

Let’s check for 795. It ends in 5, so it is divisible by 5. What about divisibility by 3?

7 + 9 + 5 = 21 ÷ 3 = 7 (divisible by 3).

Since it is divisible by both 3 and 5, 795 is divisible by 15