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Aptitude Tests - Divisibility Rules - Concept Divisibility rules are useful to f...

# Aptitude Tests - 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 a number, if the last digit is 0, 2, 4, 6, 8 i.e., divisible by 2, then the  number is divisible by 2.
Example:  123456, 208, 304…

Divisibility Rule for 3:
If the  the sum of all digits of a number is  divisible by 3 then the 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:
If the last two digits of a number are zeroes or the number formed by last two digits of  a  number  is divisible by 4 , then the  number is also divisible by 4.
Example:  31428, 44300….

Divisibility Rule for 5:
If the last digit of a number is  0 or 5 ,then the number is exactly divisible by 5.
Example:  400, 405, 1055

Divisibility Rule for 6:
If a number is exactly divisible by both 2 and 3 ,  then the  number is also divisible by ‘6’.
Example:  324 ,  1254
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 , so the number is exactly divisible   by 6.

Divisibility Rule for ‘7’:
In aptitude tests , we usually don't find the questions on divisibility by 7.. But it will be helpful to do some quicker calculations .
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’
If the number formed by  last three digits of a number is divisible by ‘8’ or last three digits of the given number are zeroes then the number is divisible by ‘8’.
Example : 56847000 , 9432256
Example:    In the number 27358128 , the number formed by last three digits is 128.
128 is  exactly divisible by 8, then the number  27358128 is exactly divisible by 8

Divisibility Rule for ‘9’:
If the sum of all digits is a number is  divisible by 9 , then the number is divisible by 9.
Example:   172845  .
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’:
In aptitude tests, usually questions on divisibility by 13 are not appeared. But here we discussed, because it useful for Faster calculations.
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

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
The remainder 147 is exactly divisible by 7 .
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 by 5.
The number 5415 is divisible by both 3 and 5 , so 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 by 5,  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.
MultiplY 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 number 3895 is also divisible by 19.

For solved problems and examples on Divisibility Rules - go through

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