Divisibility Rules-Formation & Proof from 2 to 19|Combined Rule for 7,11&13

Sohel Sahoo: Hello guys,
      In this article, I have brought an overview of divisibility rules of natural numbers & their application to solve problems in hands down.

You can also read : 

 Special divisibility Rules  

Here, I have considered a four-digit number abcd for the sake of simplicity.

This is a generalized proof of all  divisibility rules.

(Remainder method is employed to arrive at the results.)

# Proof of divisibility rules for 2,4,8,16,32..etc(powers of 2):-

Consider any number say abcd  (4 digits)

We know that, in place value decimal representation;

abcd=1000 a+100 b+10 c+d

       =10³a+10²b+10c+d

        =2³.5³a+2².5²b+2.5c+d

Divisibility test of 2:-

Now,abcd=2(2².5³a+2.5²b+5c)+d

The no. abcd is divisible by 2 iff the unit digit d is divisible by 2 or ends in 0/2/4/6/8.

    Divisibility rule of 4:-

abcd=2²(2.5³a+5²b)+10c+d
        =4(2.5³a+5²b)+cd

Clearly,the above no. is divisible by 4 iff the last two digits cd  are divisible by 4 or simply ends in 00.

     Divisibility rule of 8:-

abcd=2³(5³a)+100b+10c+d

        =8(5³a)+abc

Obviously, the no. abcd is divisible by 8 iff the last /trailing 3 digits bcd  are divisible by 8 or ends in 000.

One can proceed likewise & in the end obtain that any no. is divisible 16,32,…2ⁿ;nEN if&only if the last 4,5….n digits are divisible by them respectively,which can be generalized in this way &examined to be true.

# Proof of divisibility rule of 9&3:-

Let us Consider any number say abcd  (4 digits)

We know that, in place value decimal representation;

abcd=1000 a+100 b+10 c+d

Now, rearranging the above expansion using appropriate manipulations, we get;

abcd=999a+a+99b+b+9c+c+d

       =999a+99b+9c+(a+b+c+d)

·         Divisibility rule of 3:-

abcd=3(333a+33b+3c)+(a+b+c+d)

Here abcd will be divisible by 3 ,if the right part (a+b+c+d) is divisible by itself.

So,in general, any no. is divisible by 3 iff the sum of the digits is divisible by 3.

·         Divisibility test for 9:-

abcd=9(111a+11b+c)+(a+b+c+d)

The above no. abcd will be divisible by 9 if (a+b+c+d) is divisible by itself.

Hence,any no. is divisible by 9 iff the sum of the digits is divisible by 9.

# Proof for divisibility rule of 5&10:-

Consider any number say abcd  (4 digits)

We know that, in place value decimal representation;

abcd=1000 a+100 b+10 c+d

·         Rule for 5:-

abcd=5(200a+20b+2c)+d

This will be divisible by 5 if d is a multiple of 5.

Hence,any no. is divisible by 5 iff its unit digit ends in 0/5.

·        Divisibility rule of 10:-

abcd=10(100a+10b+c)+d

Above no. abcd will be divisible by 10 if  d is a multiple of 10.

Therefore,any no. is divisible by 10 iff it’s unit digit ends in 0 only.

# Proof of Divisibility rule of 11:-

Let ‘s consider a five digit number say abcde. 

In place value decimal representation,we have;

abcde=10000a+1000b+100c+10d+e

Rearranging the above equation using appropriate manipulations,we get;

abcde=9999a+a+1001b—b+99c+c+11d—d+e

         =(9999a+1001b+99c+11d)+(a—b+c—d+e)

          =11(99a+91b+9c+d)+(a—b+c—d+e)

So, the above no. has  got to be divisible by 11 then (a—b+c—d+e) must be divisible by 11.

i.e. a—b+c—d+e
   =(a+c+e)—(b+d) must be a multiple of 11.

Henceforth,we obtain the divisibility rule of 11 that states that;

Difference between the sum of digits in the alternate (odd & even) place has to be 0/11/any multiple of 11 iff the no. is divisible by 11.

This can be extended for any n- digit number and can be generalized in a similar procedure explained above.

# Proof of divisibility  rule of 6,12,14..etc (Composite no.s):-

Any composite can be expressed as the product of  2 (or more ) numbers (not necessarily prime) whose divisibility rule has already been known.

But we  consider that no.s which in lesser multiplied together becomes the desired no. & have individually divisibility rule independent of others as explained below.

 e.g. ,  6=2 x 3

So, any no. is divisible by 6 when it is divisible by both 2&3.

12=3 x 4 or 12= 2 x 6

Likewise,any no. is divisible by 12 if it is divisible by 3&4 as well.

But , here notice that in order to get divided by 6 we have to test the given no. for 2&3 divisibility rules and this is not sufficient enough to produce our desired output 12.

Hence, the Divisibility test of 6 depends on both divisibility tests of 2&3.

Here onwards you can check  &find yourself divisibility rules of any composite no. that has more than 2 factors. 

# Proof of divisibility  rule for 7,13,17,19…etc(prime no.s):-

Consider any number say abcd  (4 digits)

We know that, in place value decimal representation;

abcd=1000 a+100 b+10 c+d

·         Divisibility rule of 7:-

Now,by doing appropriate manipulations this can be arranged as;

abcd=700a+300a+70b+30b+7c+3c+7d—6d

      =(700a+70b+7c+7d)+(300a+30b+3c—6d)
   =7(100a+10b+c+d)+3(100a+10b+c—2d)
       =7(abc+d)+3(abc—2d)

Obviously,the above no. abcd is divisible by 7 iff (abc—2d) is divisible by 7.

Thus,any no. is divisible by 7 iff double the unit digit,then subtract the result so obtained from the rest of the digits is divisible by 7.Repeat this process if needed.

·        Divisibility rule of 13:-

In similar manner;

abcd=1000a+100b+10c+d

       =1300a—300a+130b—30b+13c—3c+13d—12d

      =(1300a+130b+13c+13d)—(300a+30b+3c+12d)

        =13(100a+10b+c+d)—3(100a+10b+c+4d)

       =13(abc+d)—3(abc+4d)

So,the above no. is divisible by 13 only if (abc+4d) is a multiple of 13.

Hence,any no. is divisible by 13 iff 4 times the last digit,then add the result from the rest of the digits is divisible by 13.Repeat this process through steps if needed. 

In this manner, you can form the Divisibility rule of 19 and onwards prime numbers.

e.g;
     1.  Check whether 92736 is divisible by 7 or not.
sol:- 92736=>9273--2 x 6=9261
                    =>926--2 x 1=924
                    =>92--2 x 4=84(You might know that 84 is a multiple of 7.So,repeat the process if required.)
                  =>8--2 x 4=0
  As 0 is divisible by 7,therefore,92736 is divisible by 7.

 2. Check whether 5499 is divisible by 13 or not.
sol:- 5499=>549+4 x 9=585
                  =>58+4 x 5=78 (You might know that 78 is a multiple of 13.So,repeat the process)
               =>7+4 x 8=39
 As 39 is divisible by 13 ,hence,5499 is also divisible by 13.

# Combined divisibility rule of 7,11&13:-

7 x 11 x 13=1001

Note:-If a number is divisible by 1001 ,then clearly divisible by 7,11&13.

                And hence also divisible by its factors viz. 77,91,143.

Consider a 6 digit no. be abcdef.

In place value decimal representation ,we have;

abcdef=abc x 1000+def

                =abc x 1001—abc+def

                =1001 x abc+(def—abc)

Hence,the above no. is divisible by 1001 i.e. 7,11&13 iff (def—abc) is divisible by them.

Let say a 9 digit no. be abcdefghi.

 abcdefghi=abc x 1000000+def x 1000+ghi

                =abc x 1001000—abc x 1000+def x 1001—def+ghi

                =abc x 1001000—abc x 1001+abc+def x 1001—def+ghi

                =1001(abc x 1000—abc+def)+(abc+ghi—def)

Therefore, the above no. is divisible by 1001 i.e.7,11&13 iff (abc+ghi—def) is divisible by them.

This can be extended for any n- digit number and can be generalized in a similar procedure explained above.

* Note:- 1.Make pairs of 3 digits from the right hand side.

                2.Take difference between sum of pairs in alternate (odd & even) positions.

                3.If that difference is divisible by 7,11&13; then the given no. is divisible by 7,11&13 respectively.

eg;  

       check whether 493948 is divisible by 7/11/13.

sol:-493948=>948--493=455

 This 455=>45--2 x 5=35;divisible by 7

       455=>(5+4)--5=4;not divisible by 11

   455=>45+4 x 5=65;divisible by 13

Therefore, the given no. 493948 is divisible by 7&13.

I will give a universal method for finding divisibility rules of any no. & a more convenient proof of combined rule for 7,11&13 by remainder method in the next article of divisibility.

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