## Divisibility rules – Number properties – Understand and apply!

In this article, we will discuss the different Divisibility rules and explain the logic behind the rules. The main numbers whose divisibility rules we would discuss are: 2, 4, 8 (and other powers of 2), 5, 25 (and other powers of 5), 3, 9, 11, and the two numbers 7 and 13 which have divisibility rules quite unlike the others.

#### Let us first enumerate the divisibility rules for 2s and 5s:

**2**^{K}**–**A number is divisible by 2^{K}when the number formed by its last ‘k’ digits is divisible by 2^{K}.**5**^{K}**–**A number is divisible by 5^{K}when the number formed by its last ‘k’ digits is divisible by 5^{K}.

Let us understand the above logic of the above divisibility rules using the number 1356. We need to check if it is divisible by 2, 5, 4, 25, 8 and 125:

- 1356 = 135 x 10 + 6: Since 10 is a multiple of 2 (and 5) we just need to verify if 6 is also a multiple of 2, which it is. Hence, the number is divisible by 2. Thus, we just need to focus on the last (units) digit of the number. Similarly, for 5, we can follow the same line of reasoning.
- 1356 = 13 x 100 + 56: Since 100 is a multiple of 4 (and 25) we just need to verify if 56 is also a multiple of 4, which it is. Hence, the number is divisible by 4. Thus, we just need to focus on the last two digits of the number. Similarly, for 25, we can follow the same line of reasoning.
- 1356 = 1 x 1000 + 356: Since 1000 is a multiple of 8 (and 125) we just need to verify if 356 is also a multiple of 8, which it is not. Hence, the number is not divisible by 8. Thus, we just need to focus on the last 3 digits of the number. Similarly, for 125, we can follow the same line of reasoning.

Note:

- For 5 , we may simply check if the last digit is 0 or 5
- For 25 , we may simply check if the last 2 digits are 00, 25, 50, or 75
- For 125 , we may simply check if the last 3 digits are 000, 125, 250, 375, 500, 625, 750, or 875

#### Let us check the divisibility rules for 3 and 9:

**3 –**A number is divisible by 3/9 when the sum of its digits is divisible by 3 respectively.**9 –**A number is divisible by 3/9 when the sum of its digits is divisible by 9 respectively.

** **Let us understand the logic of the above divisibility rules using the number 1356. We need to check if it is divisible by 3 and 9:

** **1356 = 1 x 1000 + 3 x 100 + 5 x 10 + 6

= (1 x 999 + 1) + (3 x 99 + 3) + (5 x 9 + 5) + 6

= (1 x 999) + (3 x 99) + (5 x 9) + 1 + 3 + 5 + 6

The parts within the parenthesis are multiples of 9 (and hence also 3). Thus, we just need to focus on (1 + 3 + 5 + 6), which is the sum of digits. We can check that it is not a multiple of 9 (or 3). Thus, the number is not divisible by 3 or 9.

#### Important: What are the divisibility rules for numbers like 6, 12, 72, 44, etc.?

We know that if a number N is divisible by two divisors P and Q, N is also divisible by the LCM (Least common Multiple) of P and Q, and not necessarily the product of P and Q. For example, if we have a number divisible by 4 and 6, the number is divisible by 12, not necessarily by 24 (since the LCM of 4 and 6 is 12). Different possible values of the number are 12, 24, 36, 48, etc.

Thus, if we need to check the divisibility of a number by 72, we should ideally break 72 in 2 parts which have no common factors and the LCM of which is 72. Thus, for 72, we should check the divisibility using 8 and 9. Thus, the divisibility rule for 72 is the same as the the divisibility rules for 8 and 9 considered together. Similarly, we can extend the logic to other numbers.

#### Let us check the divisibility rule for 11:

**11 – **A number is divisible by 11, if the difference between the sum of the digits in the odd places and in the even places of the number is divisible by 11.

Let us understand the above logic using the number 1356:

1356 = 1 x 10^{3} + 3 x 10^{2} + 5 x 10^{1} + 6 x 10^{0}

= 1 x (11 – 1)^{3} + 3 x (11 – 1)^{2} + 5 x (11 – 1)^{1} + 6 x 1

When we divide the above number by 11, the remainder from the

- 1
^{st}term from the right = (+1) x 6 - 2
^{nd}term from the right = (–1) x 5 - 3
^{rd}term from the right = (+1) x 3 - 4
^{th}term from the right = (–1) x 1

Thus, the total remainder = – 1 + 3 – 5 + 6 = 3; hence the number is not divisible by 11.

Thus, we start counting the digits of the number from right to left with the 1^{st}, 3^{rd}, etc. positions as odd positions and 2^{nd}, 4^{th}, etc. positions as the even positions. Thereafter, we add the digits in the odd positions and the even positions separately and subtract them.

Thus, this is the logic behind the divisibility rule of 11.

#### The divisibility rules for 7 and 13:

(Note: The discussion of divisibility rules for these two numbers is purely for academic interest)

The divisibility rules of 7 and 13 are similar to that for 11. Let us understand the above logic using the number 2141356:

2141356 = 2 x 10^{6} + 1 x 10^{5} + 4 x 10^{4} + 1 x 10^{3} + 3 x 10^{2} + 5 x 10^{1} + 6 x 10^{0}

When we divide the above number by 11, the remainder from the

- 1
^{st}term from the right = (+1) x 6 … since 1 divided by 7 leaves remainder 1 - 2
^{nd}term from the right = (+3) x 5 … since 10 divided by 7 leaves remainder 3 - 3
^{rd}term from the right = (+2) x 3 … since 100 divided by 7 leaves remainder 2 - 4
^{th}term from the right = (–1) x 1 … since 1000 divided by 7 leaves remainder –1 - 5
^{th}term from the right = (–3) x 4 … since 10000 divided by 7 leaves remainder –3 - 6
^{th}term from the right = (–2) x 1 … since 100000 divided by 7 leaves remainder –2 - 7
^{th}term from the right = (+1) x 2 … since 1000000 divided by 7 AGAIN leaves remainder 1

Thus, the total remainder = +2 – 2 – 12 – 1 + 6 + 15 + 6 = 14; hence the number is divisible by 7.

Thus, the rule for 7 is:

**7**– For 7, we have the series: {-2, -3, -1, 2, 3, 1} starting from the right end going to the left and repeating. We need to find the sum of the product of consecutive digits of the number with the numbers in the series starting from the right to left. The number is divisible by 7 if the above sum is divisible by 7.**13**– The rule is similar to that of 7, only that the series is {4, 3, -1, -4, -3, 1}.

As an exercise, try to establish the above series for 13 yourselves!

For any queries regarding the courses, feel free to talk to our mentors and our counselors. You may book a **free** counseling session at any of our education centers. Also, you may reach us at **+919433063089** and mail us at **info@cubixprep.com**. Visit our website for details about our courses and our teaching methodology.

Know what our students have to say about us. Read their success stories!

Divisibility Divisibility rules Number properties Numbers Quantitative aptitude