Pecentage Previous Year Question Given in SSC CHSL

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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      

c(NaOH) =
n(NaOH) / V(NaOH)
= 1.2345 mol dm-3

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

 2/3

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


For more solved problems and online tests, visit 
http://www.9exams.com

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