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# Divisibility Rules, Study Notes, Material – ( KVS PRT & TGT)

By BYJU'S Exam Prep

Updated on: September 13th, 2023

In this article, we should read related to the Divisibility Rules**,** Important for the KVS.

In this article, we will be covering the following topics according to the questions being asked in the KVS:-

1. Simple and Easy way to learn Divisibility Rules

**Divisibility by 2****Divisibility by 3****Divisibility by 4****Divisibility by 5****Divisibility by 6****Divisibility by 7****Divisibility by 8**

**Simple and Easy way to learn Divisibility Rules**

We now take up the interesting question as to how one can determine whether a certain given number, however large it may be, is divisible by a certainly given divisor. There is no defined general rule for checking divisibility. For different divisors, the rules differ at large. We will discuss the rule for divisors from 2 to 19.

**Divisibility by 2:**

**Rule:** *Any number, the last digit of which is rather even or zero, is divisible by 2.*

**Ex:** 12, 86 and 130 are divisible by 2 but 13, 133 and 193 are not divisible by 2.

**Divisibility by 3:**

**Rule:** *if the sum of the digits of a number is divisible by 3, the number is also divisible by 3.*

**For example:**

- 123: 1 + 2 + 3 = 6 is divisible by 3; hence 123 is also divisible by 3.
- 5673 : 5 + 6 + 7 + 3 =21; therefore divisible by 3.

**Divisibility by 4:**

**Rule:** *if the last two digits of a number are divisible by 4, the number is divisible by 4 the number having two or more zeros at the end is also divisible by 4.*

**For example:**

- 526428: 28 is divisible by 4. Therefore, the number is divisible by 4.
- 5300; there are two zeros at the end, so it is divisible by 4.

**Note:** The same rule is applicable to check the divisibility by 25. That is, a number is divisible by 25 if its last two digits are either zeros or divisible by 25.

**Divisibility by 5:**

**Rule:** *If a number ends in 5 or 0, the number is divisible by 5.*

**For example:**

- 1345: As its last digit is 5, it is divisible by 5.
- 1340: as its last digit is 0, it is divisible by 5.

**Divisibility by 6:**

**Rule:** *If a number is divisible by both 3 and 2, the number is also divisible by 6. So, for a number to be divisible by 6,*

- the number should end with an even digit or 0 and
- the sum of its digits should be divisible by 3.

**For example**:

- 63924: the first condition is fulfilled s the last digit (4) is an even number and also (6+3+9+2+4=) 24 is divisible by 3; therefore the number is divisible by 6.
- 154: The first condition is fulfilled but no the second; therefore, the number is not divisible by 6.

**Special Cases**

The rules for divisibility 7, 13, 17, 19 … are very much unique and are found very rarely. Before going on with the rule, we should know some terms like “one- more” osculator and negative osculator.

“One- more” osculator needs the number needs one more to be a multiple of 10.

**For example:**

osculator for 19 needs 1 to become 20 (=2×10), thus the osculator for 19 is 2 (taken from __2__ × 10 =20). Similarly osculator for 49 is 5 (taken from __5__ × 10 =50)

A negative osculator means the number should be reduced by one to be a multiple of 10.

**For example :**

Negative osculator for 21 is 2 (taken from __2__ × 10 = 20).

Similarly, negative osculator for 51 is 5 (taken from __5__ × 10 = 50)

**Note:** (1) What is the osculator for 7?

Now, we look for that multiple of 7 × 3 =21, as 21 is one more than __2__ × 10; our negative osculator is 2 for 7.

And 7 × 7 =49 or 49 is one less than 5 × 10; ** “one – more ” osculator** is 5 for 7.

Similarly, **osculator** for 13, 17 and 19 are ;

For 13 : 13 × 3 =39, “ one more ” osculator is 4 (from __4__ × 10)

(2) Can you define osculators for 29, 39, 21, 31,27 and 23.

(3) Can you get any osculators for an even number or a number ending with ‘5’?

**Divisibility by 7:**

First of all, we recall the osculator for 7. Once again, for your convenience, as 7 × 3 = 21 (one more than 2 ×10), our negative osculator is 2. This osculator of any number by 7. See how it works:

**Ex 1:** Is 112 divisible by 7?

**Solution:**

Step I : 11 2 : 11 – 2 × 2 = 7

As 7 is divisible by 7, the number 112 is also divisible by 7.

**Divisibility by 8:**

**Rule :**

*If the last three digits of a number is divisible by 8, the number is also divisible by 8. Also, if the last three digits of a number are zeros, the number is divisible by 8.*

Ex 1: 1256: As 256 is divisible by 8, the number is also divisible by 8.

Ex 2: 135923120: as 120 is divisible by 8, the number is also divisible by 8.

**Divisibility by 9:**

**Rule:** *If the sum of all the digits of a number is divisible by 9, the number is also divisible by 9.*

Ex 1: 39681 : 3 +9 +6+8+ =27 is divisible by 9, hence the number is also divisible by 9.

**Divisibility by 10:**

**Rule:** *any number which ends with zero is divisible by 10. There is no need to discuss this rule.*

**Divisibility by 11:**

**Rule:** *If the sums of digits at odd-even places are equal or differ by a number divisible by 11, then the number is also divisible by 11.*

**Ex 1:** 3245682 : S1 = 3+ 4 + 6+ 2 = 15 and S2 = 2+5+8 = 15

As S1 = S2, the number is divisible by 11.

**Divisibility by 12:**

**Rule:** *Any number which is divisible by both 4 and 3, is also divisible by 12.*

*To check the divisibility by 12, we*

*first divide the last two-digit number by 4. If it is not divisible by 4, the number is not divisible by 12. If it is divisible by 4 then**check whether the number is divisible by 3 or not.*

**Ex 1:** 135792: 92 is divisible by 4 and also (1+3+5+7+9+2=)27 is divisible by 3; hence the number is divisible by 12.

**Divisibility by 13:**

**Rule:** *Osculator for 13 is 4 (see note). But time, our osculator is not negative (as in the case of 7). It is ‘one-more’ osculator. So, the working principle will be different now. This can be seen in the following examples.*

**Ex 1:** Is 143 divisible by 13?

**Solution :** __14__ __3__ : 14 + 3 × 4 = 26

**Divisibility by 14:**

*Any number which is divisible by both 2 and 7, in also divisible by 14. That is, the number’s last digit should be even and at the same time, the number should be divisible by 7.*

**Divisibility by 15:**

*Any number which is divisible by both 3 and 5 is also divisible by 15.*

**Divisibility by 16:**

*Any number whose last 4 digit number is divisible by 16 is also divisible by 16.*

**Divisibility by 17:**

*The negative osculator for 17 is 5 (see note). The working for this is the same as in the case of 7.*

**Ex1:** cheek the divisible of 1904 by 17.

**Solution :**

190 4 : 190 – 5 × 4 = 170

Since 170 is divisible by 17, the given number is also divisible by 17.

Note: students are suggested not to go up to the last calculation. Whenever you find the number divisible by the given number on the right side of your calculation stop further calculation and conclude the result.

**Divisibility by 18:**

**Rule:** *Any number which is divisible by 9 and has its last digit even or zero, is divisible by 18.*

Ex1: 926568: digit -the sum is a multiple of nine (i.e. divisible by 9) and unit digit(8) is even, hence the number is divisible by 18.

**Divisibility by 19:**

*If you recall, the ‘one- more’ osculator for 19 is 2. The method is similar to that of 13, which is well known to you. Let us take an example,*

**Ex1:** 149264

**Solution :** 1 4 9 2 6 4

19 / 9 / 12 / 11 / 14

Thus, our number is divisible by 19.

**Note:** you must have understood the working principle (see the case of 13).

This article tends to be beneficial for the following exams – REET, UPTET, CTET, Super TET, DSSSB, KVS, etc.

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