Divisibility Rules

This is one of the important concept of number system .Questions on remainders and factor are quite frequent in CAT these days . Lets see the divisibility rule of few numbers.

We all perhaps know what is the divisibility rule of 2,4,6,8,3,9 .But do everyone know the logic behind it ? 

First case :

Let me start from divisibility of 2
 

If last digit is divisible by 2 , then it is divisible by 2 ...

why ?
 

Last digit is nothing but divide by 10 ...10 is perfectly divisible by 2..
 

Now we can extend this  for 4 ..

lets take last digit --> divide by 10 -->  10 /4 --> Not perfectly divisible

Now move to next digit--> divide by 100

100 is perfectly divisible by 4 .. So divisibility rule of 4 is last two digits to be divisible by 4
 

Now this goes on ... jus try for various numbers like 8 ,16,32

the above logic works for factor of 10^n  like powers of 2 and 5 . Other numbers can't be treated this way

Second case
 

Lets take an example of next number .. take 3

10/3 --> not divisible

100/3 --> not divisible

1000/3 --> not divisible and it goes on ..

But in all three cases , the remainder is 1 .

If remainder is 1 , add the digits

Divisibility rule of 3 will add the digits and then check if its divisible by 3  . This is applicable for numbers which leaves remainder 1 when 10^n is divided by that number

Quick divisibility rule :

Take 33 :

100/33 leaves remainder 1 . So one can frame divisibility rule of 33 as sum of digits taken two at a time from the right hand side of number .

Eg . Is 2112 divisible by 33 ?

12 + 21 = 33 . 33/33 perfectly divisible . Yes .. isn’t that easy ?

Third case :

Lets take an example of next number .. take 7

10/7 --> not divisible but remainder is not 0 or 1 or -1

100/7 --> not divisible but remainder is not 0 or 1 or -1

1000/7 --> not divisible but remainder is -1

so every three digit we will have -1 remainder ..

if 10^3 has -1 remainder , 10^6 has 1 as remainder...

so its alternative in nature for every three digits .

Divisibility rule of 7 will difference of sum of  the digits taken at three at a time alternatively from RHS of the number  and then check if this resultant divisible by 7

say number is 123130 to check for divisibility by 7 .,..

so according to this concept we have , 130 - 123= 7 which is divisible by 7

Or let’s say 7123130 -- > (130 + 7) – 123 = 14 which is divisible by 7

The beauty of this method is it works well for big number divisors (for instance- 143) .

235378/143 .. is it divisible ?

10^3 / 143 = remainder is -1

By this method ,we can say 378 - 235 = 143.. 143 / 143 is perfectly divisible .That's all you r done

Applying this formula , we can test divisibility of 2,3,4,5,7,8,9,11,27,13,33,37,99,77,143,259(37*7)   etc ....

Just work on these lines .. You can easily find the divisibility of all above numbers

Ps : Feel free to revert me back in comments if you have any doubts .