So many times we are asked whether a given number is divisible by some number or not. How do we check that? For smaller numbers it is easier for us to answer, but what if we are asked whether 539274921294 is divisible by a random number, say 3, or 5, or 17? How do we answer that now? It is possible to answer by applying some divisibility rules. Letâ€™s start with the basics first.

**Divisibility Rules **

A divisibility rule is a shortcut for checking whether a given number is divisible by a fixed divisor without performing the division, usually by examining its digits. Most of the divisibility rules are derived from a concept of remainders.

The two basic rules for finding out whether the number is divisible or not by any particular number:

1. If the divisor is prime number, then we can directly check for it.

2. If the divisor is a composite number then do factorization of divisor into factors, which are co-prime and check the divisibility for each prime factor individually.

**Divisibility check for 2**: Check the last digit of the given number. If it is even then the number will be divisible by 2. So if the unit digit is (0,2,4,6,8) the number will surely be divided by 2.

Why this short trick works:

We can understand this by following example: –

**Example:** Check whether 86372 is divisible by 2.

**Solution:** 86372 can be written as 80000+6000+300+70+2

Now dividing this by 2, we get:

**Example: **Check whether 2478325 is divisible by 2.

**Solution:** 2478325 can be written as 2000000+400000+70000+8000+300+20+5

**Example**: Check whether 597845657394 is divisible by 2.

**Solution:** Check for unit digit, which is 4 and is divisible by 2. So the number is divisible by 2.

**Divisibility check for 4: **Check for last 2 digits of number, if 2 digits can be divided by 4 then the number will be divided by 4**. **

**Â ****Example**: Check whether 69897665644 is divisible by 4.

**Solution:** Check for last 2 digits, 44 is divisible by 4. So the number is divisible by 4.

**Â ****How to find divisibility for any DIVISOR in form:**

**Example: **check whether 69897665644 is divisible by 8.

**Solution:**

**Example: **check whether 59374593454800 is divisible by 16.

**Solution:**

**Divisibility check forÂ 3: **Number is divisible by 3 if the sum of all the digits of number is divisible by 3.

**Example: **check whether 5376978 can be divided by 3.

**Solution: **Sum = 5+3+7+6+9+7+8=45

This number will be divisible by 3.

**Â **Why this short trick works:

**Example: **check whether 5376 can be divided by 3.

**Solution: **5376 = 5*1000+3*100+7*10+6

Now as per question:

**Divisibility check for 9: **Number is divisible by 9 if the sum of all the digits of number is divisible by 9 .

**Example: **check whether 5376978 can be divided by 9 .

**Solution: **Sum = 5+3+7+6+9+7+8=45

This number will be divisible by 9.

**Divisibility check for 27: **Number is divisible by 27 Â if the sum of block of 3 from right to left is divisible by 27 .

**Example: **check whether 828279 can be divided by 27.

**Solution: **Sum = 828 + 279 =1107

This number will be divisible by 27.

**How to find divisibility for any DIVISOR in Â form: **

**Example: **check whether 69897665625Â is divisible by 25 Â .

**Solution: **

Step 1 :- Â 25 is of Â form.

Step 2 :- Â n=2Â so we will consider last 2 digits of the number.

Step 3 :-Â so last 2 digits should be divisible by .

In this case 25 is divisible by divisor 25. So the number will be divisible by 25.

**NOTE**: Â Last 2 digits to be divided 25 should be (00,25,50 or 75 )

**Â ****Example: **check whether 94382937625 is divisible by 125.

**Solution:**

Step 1 :-Â 125 is of Â form .

Step 2:- Â n=3Â so we will consider last 3 digits of the number .

Step 3:-Â so last 3 digits should be divisible by In this case 625 is divisible by divisor 125. So the number will be divisible by 125.

Letâ€™s practice some more questions.

**EXERCISE:**

**Question 1: **Check divisibility of 124 by 2, 3, 4 and 5

### Answer and Explanation

**Solution:**

124 is divisible by 2 since the last digit is 4.

124 is not divisible by 3 since the sum of the digits is 7 (1+2+4 = 7), and 7 is not divisible by 3.

124 is divisible by 4 since 24 is divisible by 4.

124 is not divisible by 5 since the last digit is 4 it is neither 0 nor 5.

124 is divisible by 2 and 4 and not by 3 and 5.

**Question 2: **If 5432*7 is divisible by 9, then the digit in place of * is:

(1) 0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (2) 1Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (3) 6Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (4) 9

(SSC Combined Graduate Level Prelim Exam. 04.07.1999 (Second Sitting)

### Answer and Explanation

**Solution (3)**

A number is divisible by 9, if sum of its digits is divisible by 9.

Let the number b x.

5 + 4 + 3 + 2 + x + 7 = 21 + x

x= 6. (as x cannot be two digit)

**Question 3: **The least number, which must be added to 6709 to make it exactly divisible by 9, is

(1) 5Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (2) 4Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (3) 7Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (4) 2

(SSC Combined Graduate Level Prelim Exam. 08.02.2004 (First Sitting)

### Answer and Explanation

**Solution (1)**

A number is divisible by 9 if the sum of its digits is divisible by 9

Here, 6 + 7 + 0 + 9 = 22.

Now, 22 + 5 = 27, which is divisible 9. Hence 5 must be added to 6709.