Lesson Divisibility by 9 rule

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Divisibility by 9 rule


The  "Divisibility by 9" rule  is as follows:


    An integer number is divisible by  9  if and only if the sum of its digits is divisible by  9.    


In other words,  for checking if the given integer number is divisible by  9,  make the following steps:

    1.  Calculate the sum of the digits of the given number.

    2.  Check if this sum is divisible by  9.

    3.  If the sum of the digits is divisible by  9,  then the original number is divisible by  9.
         If the sum of the digits is not divisible by  9,  then the original number is not divisible by  9.

Quite simple,  isn't?


Example 1

Check if the number  576  is divisible by  3.

Solution

The sum of the digits of the given number is  5 + 7 + 6 = 18.
It is divisible by  9.
Hence,  the original number  576  is divisible by  9,  in accordance with the  "Divisibility by 9" rule.

You may check it by making the direct division:  576%2F9 = 64.  It shows that the number  376  is divisible by  9.
The  Divisibility rule  allows you to get the same conclusion without making long calculations.


Example 2

Check if the number  728  is divisible by  9.

Solution

The sum of the digits of the given number is  7 + 2 + 8 = 17.
It is not divisible by  9.
Hence,  the original number  728  is not divisible by  9,  in accordance with the  "Divisibility by 9" rule.


Below are even more impressive examples.


Example 3

Check if the number  777,777,777  is divisible by  9.

Solution

The sum of the digits of the given number is  7 + 7 + 7 + 7 + 7 + 7 + 7 + 7 + 7 = 63.
It is divisible by  9.
Hence,  the original number  777,777,777  is divisible by  9,  in accordance with the  "Divisibility by 9" rule.

Again, the  Divisibility rule  allows you to get the conclusion without making long calculations.


Example 4

Check if the number  7,777,777,777  is divisible by  9.

Solution

The sum of the digits of the given number is  7 + 7 + 7 + 7 + 7 + 7 + 7 + 7 + 7 + 7 = 70.
It is not divisible by  9.
Hence,  the original number  7,777,777,777  is not divisible by  9,  in accordance with the  "Divisibility by 9" rule.


The proof of this  "divisibility rule"  is exceptionally funny.
It is very close to the proof of the  "divisibility by 3" rule  in the lesson  Divisibility by 3 rule  under the current topic in this site.

The proof of the "divisibility by 9" rule


First,  let us consider how the  "divisibility by 9"  rule can be proved for an arbitrary three-digit number  "abc",  for example.

Our three-digit number  "abc"  is   100*a + 10*b + c.   (It is how our decimal system works).

Now, regroup it:   100*a + 10*b + c = (99*a+a) + (9*b+b) + c = (99*a + 9*b) + (a + b + c).

The additive  (99*a + 9*b)  is divisible by  9.  Therefore,  the divisibility by  9  of our original number  "abc"  depends on and is determined solely by the last additive  (a + b + c),  which is the sum of the digits of the number  "abc".


For the general case,  the proof follows the same logical pattern.
Let  a%5Bn%5Da%5Bn-1%5D . . . a%5B2%5Da%5B1%5Da%5B0%5D  be the decimal record of our integer number  N.
Then  N = a%5Bn%5D%2A10%5En + a%5Bn-1%5D%2A10%5E%28n-1%29 + . . . + a%5B2%5D%2A10%5E2 + a%5B1%5D%2A10 + a%5B0%5D.

Let us rewrite it in the form  N = [a%5Bn%5D%2A%2810%5En+-1%29%2Ba%5Bn%5D] + [a%5Bn-1%5D%2A%2810%5E%28n-1%29+-1%29%2Ba%5Bn-1%5D] + . . . + [a%5B2%5D%2A%2810%5E2-1%29%2Ba%5B2%5D] + [a%5B1%5D%2A%2810-1%29%2Ba%5B1%5D] + a%5B0%5D

and then regroup it as  N = a%5Bn%5D%2A%2810%5En+-1%29 + a%5Bn-1%5D%2A%2810%5E%28n-1%29-1%29 + . . . + a%5B2%5D%2A%2810%5E2-1%29-1 + a%5B1%5D%2A%2810-1%29 + [a%5Bn%5D + a%5Bn-1%5D + . . . a%5B2%5D + a%5B1%5D + a%5B0%5D].

The first  n  additives,  a%5Bn%5D%2A%2810%5En-1%29,  a%5Bn-1%5D%2A%2810%5E%28n-1%29-1%29, . . . ,  a%5B2%5D%2A%2810%5E2-1%29,  a%5B1%5D%2A%2810-1%29  are divisible by  9,  because each has a multiplier  10%5Ek-1  which is divisible by  9.  (Indeed, each multiplier  10%5Ek-1  is the number which is written with the only digit  "9"  in all positions).
Hence,  the sum of these additives  a%5Bn%5D%2A%2810%5En-1%29 + a%5Bn-1%5D%2A%2810%5E%28n-1%29-1%29 + . . . + a%5B2%5D%2A%2810%5E2-1%29 + a%5B1%5D%2A%2810-1%29  is divisible by  9  too.
Therefore,  the divisibility by  9  of our original number depends on and is determined solely by the sum  a%5Bn%5D + a%5Bn-1%5D + . . . a%5B2%5D + a%5B1%5D + a%5B0%5D,  which is the sum of all digits of the original number.   It is what has to be proved.


My other lessons in this site on divisibility rules are

    - Divisibility by 2 rule
    - Divisibility by 3 rule
    - Divisibility by 4 rule
    - Divisibility by 5 rule
    - Divisibility by 6 rule
    - Divisibility by 10 rule
    - Divisibility by 11 rule
    - Restore the omitted digit in a number in a way that the number is divisible by 9
    - Restore the omitted digit in a number in a way that the number is divisible by 11
    - Can there be a perfect square ?
    - Math circle level problems on divisibility numbers
    - Math circle level problem on restoring digit in the product of two 16-digit numbers
    - Math circle level problem on finding remainders
    - OVERVIEW of Divisibility rules by 2, 3, 4, 5, 6, 9, 10 and 11

Lessons that are closely adjacent to these are

    - Product of two consecutive integers is divisible by 2
    - Product of three consecutive integers is divisible by 6
    - Problems dealing with the product of two consecutive integers
    - Problems dealing with the product of three consecutive integers

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