Divisibility Rule for 2, 3, 4, 5, 6, 7, 8, 9, 10, 11

Divisibility Rule


Divisibility Rule for 2 3 4 5 6 7 8 9 10 11

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How to test number is completely divisible by 2, 3, 4, 5, 6, 7, 8, 9, 10, 11

Rules to divide by 2

If the unit digit of a number is any of 0, 2, 4, 6, 8 then that number will be divided by 2.

Example- Each of the following numbers will be completely divisible by 2.

687612, 987598, 96584, 5896, 67310



Rules to divide by 3

If the sum of all digits of a given number is completely divisible by 3, then that number will be completely divisible by 3.

Example- (i) 57351

Sum of digits= (5+7+3+5+1) = 21, which completely divisible by 3.

So, the given number is completely divisible by 3

Rules to divide by 9

If the sum of all digits of a given number is completely divisible by 9, then that number will be completely divisible by 9.

Example- (i) 451827

Sum of digits= (4+5+1+8+2+7) = 27, which completely divisible by 9.

So, the given number is completely divisible by 9.

➡ 9(Odd power), when divided by 10, always gives 9 as the remainder.

➡ 9(Even power), when divided by 10, always gives 1 as the remainder.





Rules to divide by 5

If the given number unit digit is 5 Or 0, then that number will be completely divisible by 5.

Example- (i) 451825 and 68930 are completely divisible by 5.

Rules to divide by 10

If the given number unit digit is  0, then that number will be completely divisible by 10.

Example- 56410, and 65980 are completely divisible by 10.





Rules to divide by 4

A given number will be divisible by 4 only when the number formed by its tens and unit digits (Last to digits) is completely divisible by 4.

Example- (i) 576832

The number formed by its tens and unit digits is 32, which will be completely divided by 4.

Hence the given number 576832 will be completely divisible by 4.


Rules to divide by 8


A given number will be divisible by 8 only when the number formed by its unit, tens and hundred digits (Last three digits) is completely divisible by 8.

Example- (i) 654512

The number formed by its tens and unit digits is 512, which will be completely divided by 8.

Hence the given number 654512 will be completely divisible by 8.

Rules to divide by 6

A given number will be divisible by 6 only if it is divisible by 2 and 3.

Example- 8976, 65136
Both numbers are divisible by 2 and 3.

Rules to divide by 7

The number is obtained by doubling the unit digit and subtracting it from the remaining number, if it is divisible by 7, then the given number becomes divisible by 7.

Example- (i) 882

(88 - 2 ✕ 2) = 84 which completely divisible by 7.
Hence the given number 882 will be completely divisible by 7.

(ii) 663

(66 - 2 ✕ 3) = 60 which completely divisible by 7.
Hence the given number 663 will be completely divisible by 7.



Rules to divide by 11

The given number will be divisible by 11 only if the difference between the sum of the digits of the even-places and the sum of the digits of the odd-places is 0 or a multiple of 11.

Example- (i) 901219

( sum of the digits of the odd-places) - (sum of the digits of the even-places) 

= ( 9+2+0 ) - (1+1+9 ) 
= (11 - 11 ) = 0
Hence the given number 901219 will be completely divisible by 11.


(ii) 82735213

(sum of the digits of the even-places) - ( sum of the digits of the odd-places)

= ( 1+5+7+8 ) - ( 3+2+3+2 )
= ( 21 - 10 )
= 11
Hence the difference is 11, which is completely divisible by 11 then the given number 82735213 will be completely divisible by 11.