Divisibility Rules - Concept
Divisibility rules are useful to find whether a number is exactly
divisible by another number or not without performing actual division.
Divisibility Rule For 0 : Division
by zero is not defined
Divisibility Rule For 1 : Every
number is divisible by 1
Divisibility Rule for 2:
In the given number, if the last digit is 0, 2, 4, 6, 8 i.e.,
divisible by 2, then the given number is divisible by 2.
Example: 123456,
208, 304…
Divisibility Rule for 3:
In the given number the sum of all digits is divisible by 3 . The
given number is also divisible by 3.
Example :- 1731
In the number 1731, sum of the digits is 1+7+3+1=12, which is
exactly divisible by 3.So 1731 is also divisible by 3.
Example: 825, 1362, 1233…
Divisibility Rule for 4:
In the given number the last two digits is divisible by 4 the
given number is also divisible by 4.
Example: 31428, 44300….
Divisibility Rule for 5:
In the given number the last digit is 0 or 5 ,then the number is
exactly divisible by 5.
Example: 400, 405, 1055…
Divisibility Rule for 6:
If the number is exactly divisible by both 2 and 3 then the given
number is also divisible by ‘6’.
Example: 324 , 1254
If we take the number 216,
The given number is even number, so it is divisible
by 2
And sum of the digits is 2+1+6 =9 , which is multiple
of 3.
So the number 216 is multiple of both 2 and 3 then the number is
exactly divisible by 6.
Divisibility Rule for ‘7’:
Take the
last digit of the number, multiply it by 2 , and subtract the
product from the rest of the number. If the answer is divisible by 7
(including 0), then the number is also divisible by 7. . Continue
this until you get a one-digit number. The result is 7, 0, or -7, if and only
if the original number is a multiple of 7.
Example: 371
371×2
-2
------
35
35 is divisible by 7. The given number 371 is also divisible
by 7
Divisibility Rule for ‘8’:
In the given number the last three digits is divisible by ‘8’ or
last three digits of the given number are zeroes then the number is divisible
by ‘8’.
Example: In the number 27358128 , the last digits 128 exactly
divisible by 8, then the number 27358128 is exactly divisible by 8
Divisibility Rule for ‘9’:
In the given number the sum of all digits is divisible by 9 then
the number is divisible by 9.
Example: 172845 .
In this number sum of the digits 1+7+2+8+4+5 =27 which is a
multiple of 9
So the number 172845 is also divisible by 9.
Divisibility Rule for 10:
In the given number last digit is ‘0’ then the number is divisible
by 10.
Example: 1000, 2120…
Divisibility Rule for ‘11’:
A number
is divisible by 11. If the difference of the sum of its digits at odd places
and sum of its digits at even places is either 0 or a number divisible by 11.
Example 1 : 14641 ,
(Sum of digits at odd places) – (sum
of digits at even places)
( 1 + 6 + 1 ) - (
4 + 4 ) = 0
The difference is 0, the number 14641 is exactly divisible by 11
Example 2 :
The number 4832718 is divisible by 11 , Since
(Sum of digits at odd places) – (sum of digits at even places)
( 8 + 7 +
3 + 4 ) - ( 1 + 2 + 8 ) = 11 which is divisible by 11.
Divisibility Rule for ‘12’:
If a number is exactly divisible by both 3 and 4 , then
the number is divisible by 12
Example:
1752
In this number, the sum of the digits is 1+7+5+2=15 . 15 is
multiple of 3, so the number is multiple of 3
1752, last 2 digits 52 exactly divisible by 4, so the number is
divisible by 4
So the number 1752 is divisible by both 3 and 4 , the number
is divisible by 12
Divisibility Rule for ‘13’:
Take the last digit of the number, multiply it by 4 ,
and add the product to the rest of the number. If the answer is
divisible by 13, then the number is also divisible by 13 . .
Continue this until you get a two-digit number. The result is multiple of 13,
if and only if the original number is a multiple of 13.
Example: 1391 => 1391 x 4
+ 4
143
As 143 is multiple of 13, the number 1391 is multiple of 13
$\frac{2000000}{30000}$
Divisibility Rule for ‘14’:
If a number is exactly divisible by both 2 and 7, then
number is exactly divisible by 14
Example: 1512 , The number
is even , so it is divisible by 2
Check it by 7, 1512*2
-4
147
147, which is multiple of 14.
The given number 1512 is exactly divisible by both 2
and 7 , so the number is divisible by 7
Divisibility Rule for ‘15’:
If a number multiple of both 3 and 5, the number is multiple of 15
Example: 5415
In this number, the sum of the digits is 5+4+1+5=15 . 15 is
multiple of 3, so the number is multiple of 3
5415, last digit is 5, so the number is divisible by5.
So the number 5415 is divisible by both 3 and 5 , the
number is divisible by 15
Divisibility Rule for ‘16’:
If last 4 digits of a number are zeroes ,or multiple of 16, then
number is multiple of 16
Example 1: 725680000 Here
last 4 digits are 0's, so the number is divisible by 16.
Example 2 : :
7293181680 , In this number last 4 digits are 1680, is a multiple of
16.So the number is multiple of 16.
Divisibility Rule for 17
Multiply the last of digit of the given number
and then subtract the product from remaining truncated number. Repeat the step
as necessary. If the result is divisible by 17, the original number is also
divisible by 17
Example : Take the
number 1938.
193 8 x5
-40
----
153.
Since 153 is divisible by 17, the original number 2278 is also
divisible.
Divisibility Rule for 18.
If a number is divisible by both 2 and 9, then
the number is exactly divisible by 18
Example : 23526
23526 is an even number, so it is divisible by 2
Sum of the digits of the number 23526 is ( 2+3+5+2+6)=18 which is
divisible by 9.
So 23526 is exactly divisible by both 2 and 9, the number is
divisible by 18.
Divisibility Rule for 19.
Multiple the last digit of the original number
by 2 and then add the product to the remaining truncated number. Repeat the
step as necessary. If the result is divisible by 19, the original number is
also exactly divisible by 19
Example : Let us check for 3895::
389 5x2
+10
-----
39 9x2
+18
-----
57
Since 57 is divisible by
19, original no. 3895 is also divisible by 19.
For solved problems and examples on Divisibility Rules - go
through Divisibility
Rules - Solved Problems
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