![divisibility.png](https://steemitimages.com/DQmVurAwKdjM7PBfhH5zGkCDHeTG2sBRQHYHMjRiSJgnJrD/divisibility.png)

<b>Checking divisibility of a number by 3,4,5,6,8,9 is  easy, but the tricky one is 7.</b>
Let's talk about the easier ones :)

<b>Divisibility by 3</b>: many of you may already know, if the sum of  all the digits of a number is divisible by  3, then the original number is divisible by 3 as well! 

   <b>Example</b>: given 369, 3+6+9 = 18 and it is divisible by 3, hence 369 is also divisible by 3!

<b>Divisibility by 4</b>: This one is quite trivial, if the number formed by the last two digits is divisible by 4., then the number is also divisible by 4 as well!

   <b>Example</b>: given 2016, 16 is divisible by 4, hence 2016 is also divisible by 4!

<b>Divisibility by 5</b>: This one is more trivial, if the last digit of the number is 0 or 5, then the number is divisible by 5:)

   <b>Example</b>: given 1020, the last digit is 0, hence 1020 is divisible by 5~

<b>Divisibility by 6</b>: Here comes a compound number, basically the follow rules works for all compound numbers:)

for a compound number, find its factors, in this case 6 = 2x3, and then you can check if the number is divisible by all of its factors.

   <b>Example</b>: given 576, 5+7+6 = 18 and it is divisible by 3, 576 itself is also divisible by 2, then 576 is divisible by 6!

<b>Divisibility by 8</b>: This one is similar to divisibility by 4, if the number formed by the last three digits is divisible by 8., then the number is also divisible by 8 as well~

   <b>Example</b>: given 10200, 200 is divisible by 8, hence 10200 is also divisible by 8!

<b>Divisibility by 9</b>: Although 9 is a compound number, but its factor contains multiple copies of themselves, i.e. 9 = 3x3, so we need a different method rather then just checking if the number is divisible by 3 two times (that's really dumb). But surprisingly, the rule is similar to 3's one. If the sum of  all the digits of a number is divisible by 9, then the original number is divisible by 9 as well.......lol

   <b>Example</b>: given 4689, 4+6+8+9 = 27 and it is divisible by 9, hence 4689 is also divisible by 9!

<b>For those who stayed till here,
FInally here is the tougher one - '7',
There will also be explanations, please read though :) </b>

Given 2023,
1. Double the last digit of the number,
In this case 3x2 = 6
2. Subtract the number from number formed by the rest of the digits,
i.e. 202 - 6 = 196
3. If the result is divisible by 7, then the original number is also divisible by 7! If you are not sure, just repeat 1&2 :)
i.e. 19 - (6x2) = 7
Congratulations :) 2023 is divisible by 7~~~what a happy ending :D


Oh ...wait........that worked....in a strange way,
but <b>WHY????</b>

Here is a more technical explanation for math nerd :)
<b> -------------------------------- MATH WARNING --------------------------------</b>

By the fact that (mod 7), 1->1, 10->3, 100->2, 1000-> 6...etc and 10x+y->10x+y-21y-> 10(x-2y)
Hence, 10x + y is divisible by 7 iff x − 2y is divisible by 7.
By subtracting the last digit twice from the number formed by the other digits, we can reduce a number's size without altering its divisibility. By iterating such a process, we will finally obtain a small number which we can tell its divisibility easily. That's how this strange method worked.