Simple tips for divisibility Technics

                                                  Checking of the Divisibility of number by the smallest prime numbers is facilitate the prime factoring any number and further uses to find out square roots and cube roots of the numbers easily.These are use full for upper class mathematics in geometry, mensuration etc.

1. Divisible by 2

                               A number can be divisible by 2, if the number is even number or the number has            even integer in it,s 1's place.
         e.g: 8546794 is divisible by 2 as 4 in the 1's place is divisible by 2,hence the entire number is              divisible by 2 exactly.

           Divisible by4
                                 a number can be divisible by 4,if the number has minimum of  first two digits                forming a number can be divisible by 4 [4, 8 are also divisible by 4], the entire number can be            divisible by 4 exactly [i.e with reminder 0] This Technic is used for checking the number                    having the number of digits>2
         E.g: let us check the number 5236 is divisible by 4 or not
        simply check the first 2 digit number 36
        36/4=9 remainder=0 hence the total number can be divisible by 4
         Justification 5236/4=1309 with reminder=0

1. Divisible by8
2.                        a number can be divisible by 4,if the number has minimum of  first three digits forming a number can be divisible by 8 [the numbers having number of places<3 can be checked by direct method], the entire number can be divisible by 8 exactly [i.e with reminder 0] This Technic is used for checking the number having the number of places>3 only.
3. Eg: let us check the number 37952128 is divisible by 8 or not
           simply check the first 3 digit number 128
   128/8=16 remainder=0 hence the total number can be divisible by 8
    Justification 37953128/8=4744141 with reminder=0

1. Divisible by3

                             A number can be divisible 3, if the total digits of the number is divisible by 3. If the           sum of the digits is more than 1 places can be added the digits further to single digit such that             the number can check easily. The remainder become the same as the reminder of the main                   number divided by 3.

        e.g number=825

       the sum of the digits=8+2+5=15, further 1+5=6,6 is divisible by 3 so the number 825 is divisible       by 3

         i.g number=1735, 

         the sum of the digits=1+7+3+5=16 further 1+6=7 is can't be divisible by 3, as the remainder=1

         the reminder is same as the reminder by main divided by 3 

• Divisible by9    
• A number can be divisible 9, if the total digits of the number is divisible by 9. If the sum of the digits is more than 1 places can be added the digits further to single digit such that the number can check easily. The remainder become the same as the reminder of the main number divided by 9.

1. e.g number=825

   the sum of the digits=8+2+5=15, further 1+5=6, 6 is not divisible by 9 so the number 825 is not divisible by 9. the reminder is 6.
   e.g number=173511 
   the sum of the digits=1+7+3+5+1+1=18 further 1+8=9, 9 can be divisible by 9,

           Divisible by 5 

          A number cab be divisible by 5, if the number has 5 or 0 in it,s 1,s place.

        e.g 5, 15, 254589731525, 125462189510 are divisible by 5.

           Divisible by 10

                                  A number cab be divisible by 10, if the number has  0 in it,s 1,s place.

        e.g 10, 150, 2545897315250, 125462189510 are divisible by 10

           Divisible by 7

                              A number cab be divisible by 7. If a number has the digits it,s places, such that the            difference between twice of the 1's place or 1st digit to the number formed remaining digits                should be divisible by 7.

           e.g is 63 divisible by 7 or not

           difference between twice the 1's place number to remaining number=2x3-6=0,0 can be                        divisible 7 hence the number can be divisible by 7

           e.g: is the number 51513 divisible by 7

           step1: 5151-2x3=5145

           step2; 514-2x5=504

           Step3: 50-2x4=42

           step4:  4-2x2=0, 0 can be divisible by 7, so the entire  number can be divisible by 7.

           If the number has more places the same process can be repeated until getting single digit.

           Divisible by 11

                                 A number can be divisible by 11, if the number has digits such that the                           difference between sums of the alternate digits is divisible by 11.

          E.g: 1059531

          1st. sum of Alternate numbers=1+5+5+1=12

          2nd sum of Alternate numbers=0+9+3=12

          Difference= 12-12=0, 0 is divisible by 11 so the entire number divisible by 11

Some example problems for competitive examinations: 

1]  Find the missing digit in the number 1?1 if the number is exactly divisible by 3
     let ?=x
     The sum of the digits in the number should be divisible by 3
     Sum of the digits=1+x+1
                           =x+2
The least number >2 can be divisible by 3 is 3
                           =>=x+2=3
                              x=3-2=1
So the missing number=1

2] Find the missing digit in the number 53?2 if the number is exactly divisible by 3
    let ?=x
    The sum of the digits in the number should be divisible by 3
     Sum of the digits=5+3+x+2
                           =x+10
   The least number >10 can be divisible by 3 is 12.
                           =>x+10=12
                                x=12-10=2
   So the missing number=2

3] Find the missing digit in the number 53?2 if the number is exactly divisible by 9
    let ?=x
    The sum of the digits in the number should be divisible by 3
     Sum of the digits=5+3+x+2
                           =x+10
  The least number >10 can be divisible by 9 is 18.
                           =>x+10=18
                                x=18-10=8
   So the missing number=8
 or again add the digits obtained in the total 10=1+0=1
  The least number >10 can be divisible by 9 is 9.
so                      x+1=9
                      => x=9-1=8.
the missing number=8

4] Find the missing digit in the number 53?2 if the number is exactly divisible by 11
    let ?=x
The difference among sum of  alternate  digits must be divisible by 11
1st alternate sum of digits=5+x
2nd alternate sum of digits=3+2=5
difference=5+x-5=x
The least  number can be divisible by 11=0
Hence the difference may equal to zero
hence x=0
So the missing number=0
Finally the number=5302.