Hi,

You all must have used divisibility rules in one of Those questions. But have you ever wondered, why is it that you take the units digit as even to check the divisibility by 2, why sum of all the digits is checked for divisibility by 3...

Still wondering, then rest of the post is for you:

Lets start with a number abc
it can be written as 100a + 10b + c, where, a, b & c are non negative integers:

Divisibility rule for 2

\frac {100a + 10b + c}2
100a, 10b are divisible by 2, so, whether the number is divisible by 2 depends on c and that's the unit digit.

Divisibility rule for 3

\frac {100a + 10b + c}3
100a+10b+c=99a+a+9b+b+c=(99a+9b)+(a+b+c)
(99a+9b) is divisible by 3, so, the number is divisible by 3 when a+b+c is divisible by 3 and that's sum of all digit in a number.

Divisibility rule for 4

\frac {100a + 10b + c}4
100a is divisible by 4, so, whether the number is divisible by 4 depends on 10b+c ,i.e, last two digits of a number.

Divisibility rule for 5

This one is similar to divisibility rule for 2.
\frac {100a + 10b + c}5, since 100a, 10b are divisible by 5, what remains is c,
so, c has to be 0 or 5 for number to be divisible by 5.

Divisibility rule for 6

The classical rule tells us that if the number is divisible by 2 & 3 it is divisible by 6.

Divisibility rule for 7

That's a tricky one,
\frac {100a + 10b + c}7
100a+10b+c=98a+2a+7b+3b+c=(98a+7b)+(2a+3b+c)
(98a+7b) is divisible by 7, so you have to only check (2a+3b+c)
For ex - 143, 2*1+3*4+1*3=17, not divisible by 7
154 = 2*1+3*5+1*4=21, divisible by 7.
For numbers with 4 or more digits, I believe you can find out the rule now!

Divisibility rule for 8

Similar to that of 2 & 4.

Divisibility rule for 9

Similar to that of 3.

Divisibility rule for 11

\frac {100a+10b+c}{11}
100a+10b+c=99a+a+11b-b+c=(99a+11b)+(a+c-b)
(99a+11b) is divisible by 11, so we have to check (a+c-b), i.e., difference of sum of alternate digits in a number.

I believe the concept behind the divisibility rule is clear to you now!

Regards,
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