Lesson Divisibility by 9 rule

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<H2>Divisibility by 9 rule</H2>

The &nbsp;"Divisibility by 9" rule&nbsp; is as follows:

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

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

&nbsp;&nbsp;&nbsp;&nbsp;1. &nbsp;Calculate the sum of the digits of the given number.

&nbsp;&nbsp;&nbsp;&nbsp;2. &nbsp;Check if this sum is divisible by &nbsp;9.

&nbsp;&nbsp;&nbsp;&nbsp;3. &nbsp;If the sum of the digits is divisible by &nbsp;9, &nbsp;then the original number is divisible by &nbsp;9.

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;If the sum of the digits is not divisible by &nbsp;9,&nbsp; then the original number is not divisible by &nbsp;9.

Quite simple, &nbsp;isn't?

<H3>Example 1</H3>Check if the number &nbsp;576&nbsp; is divisible by &nbsp;3.

Solution

The sum of the digits of the given number is &nbsp;5 + 7 + 6 = 18.

It is divisible by &nbsp;9.

Hence, &nbsp;the original number &nbsp;576&nbsp; is divisible by &nbsp;9, &nbsp;in accordance with the &nbsp;"Divisibility by 9" rule.

You may check it by making the direct division: &nbsp;{{{576/9}}} = {{{64}}}. &nbsp;It shows that the number &nbsp;376&nbsp; is divisible by &nbsp;9.

The &nbsp;Divisibility rule&nbsp; allows you to get the same conclusion without making long calculations.

<H3>Example 2</H3>Check if the number &nbsp;728&nbsp; is divisible by &nbsp;9.

Solution

The sum of the digits of the given number is &nbsp;7 + 2 + 8 = 17.

It is not divisible by &nbsp;9.

Hence, &nbsp;the original number &nbsp;728&nbsp; is not divisible by &nbsp;9, &nbsp;in accordance with the &nbsp;"Divisibility by 9" rule.

Below are even more impressive examples.

<H3>Example 3</H3>Check if the number &nbsp;777,777,777&nbsp; is divisible by &nbsp;9.

Solution

The sum of the digits of the given number is &nbsp;7 + 7 + 7 + 7 + 7 + 7 + 7 + 7 + 7 = 63.

It is divisible by &nbsp;9.

Hence, &nbsp;the original number &nbsp;777,777,777&nbsp; is divisible by &nbsp;9, &nbsp;in accordance with the &nbsp;"Divisibility by 9" rule.

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

<H3>Example 4</H3>Check if the number &nbsp;7,777,777,777&nbsp; is divisible by &nbsp;9.

Solution

The sum of the digits of the given number is &nbsp;7 + 7 + 7 + 7 + 7 + 7 + 7 + 7 + 7 + 7 = 70.

It is not divisible by &nbsp;9.

Hence, &nbsp;the original number &nbsp;7,777,777,777&nbsp; is not divisible by &nbsp;9, &nbsp;in accordance with the &nbsp;"Divisibility by 9" rule.

The proof of this &nbsp;"divisibility rule"&nbsp; is exceptionally funny.

It is very close to the proof of the &nbsp;"divisibility by 3" rule&nbsp; in the lesson &nbsp;<A HREF=http://www.algebra.com/algebra/homework/divisibility/lessons/Divisibility-by-3-rule.lesson>Divisibility by 3 rule</A>&nbsp; under the current topic in this site.

<H3>The proof of the "divisibility by 9" rule</H3>

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

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

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

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

For the general case, &nbsp;the proof follows the same logical pattern.

Let &nbsp;{{{a[n]}}}{{{a[n-1]}}} . . . {{{a[2]}}}{{{a[1]}}}{{{a[0]}}}&nbsp; be the decimal record of our integer number &nbsp;N.

Then &nbsp;N = {{{a[n]*10^n}}} + {{{a[n-1]*10^(n-1)}}} + . . . + {{{a[2]*10^2}}} + {{{a[1]*10}}} + {{{a[0]}}}.

Let us rewrite it in the form &nbsp;N = [{{{a[n]*(10^n -1)+a[n]}}}] + [{{{a[n-1]*(10^(n-1) -1)+a[n-1]}}}] + . . . + [{{{a[2]*(10^2-1)+a[2]}}}] + [{{{a[1]*(10-1)+a[1]}}}] + {{{a[0]}}}

and then regroup it as &nbsp;N = {{{a[n]*(10^n -1)}}} + {{{a[n-1]*(10^(n-1)-1)}}} + . . . + {{{a[2]*(10^2-1)-1}}} + {{{a[1]*(10-1)}}} + [{{{a[n]}}} + {{{a[n-1]}}} + . . . {{{a[2]}}} + {{{a[1]}}} + {{{a[0]}}}].

The first &nbsp;n&nbsp; additives, &nbsp;{{{a[n]*(10^n-1)}}}, &nbsp;{{{a[n-1]*(10^(n-1)-1)}}}, . . . , &nbsp;{{{a[2]*(10^2-1)}}}, &nbsp;{{{a[1]*(10-1)}}}&nbsp; are divisible by &nbsp;9, &nbsp;because each has a multiplier &nbsp;{{{10^k-1}}}&nbsp; which is divisible by &nbsp;9. &nbsp;(Indeed, each multiplier &nbsp;{{{10^k-1}}}&nbsp; is the number which is written with the only digit &nbsp"9"&nbsp; in all positions).

Hence, &nbsp;the sum of these additives &nbsp;{{{a[n]*(10^n-1)}}} + {{{a[n-1]*(10^(n-1)-1)}}} + . . . + {{{a[2]*(10^2-1)}}} + {{{a[1]*(10-1)}}}&nbsp; is divisible by &nbsp;9&nbsp too.

Therefore, &nbsp;the divisibility by &nbsp;9&nbsp; of our original number depends on and is determined solely by the sum &nbsp;{{{a[n]}}} + {{{a[n-1]}}} + . . . {{{a[2]}}} + {{{a[1]}}} + {{{a[0]}}}, &nbsp;which is the sum of all digits of the original number. &nbsp;&nbsp;It is what has to be proved.

My other lessons in this site on divisibility rules are

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/divisibility/lessons/Divisibility-by-2-rule.lesson>Divisibility by 2 rule</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/divisibility/lessons/Divisibility-by-3-rule.lesson>Divisibility by 3 rule</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/divisibility/lessons/Divisibility-by-4-rule.lesson>Divisibility by 4 rule</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/divisibility/lessons/Divisibility-by-5-rule.lesson>Divisibility by 5 rule</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/divisibility/lessons/Divisibility-by-6-rule.lesson>Divisibility by 6 rule</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/divisibility/lessons/Divisibility-by-10-rule.lesson>Divisibility by 10 rule</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/divisibility/lessons/Divisibility-by-11-rule.lesson>Divisibility by 11 rule</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/divisibility/lessons/Restore-the-omitted-digit-in-a-number-in-a-way-the-number-is-divisible-by-9.lesson>Restore the omitted digit in a number in a way that the number is divisible by 9</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/divisibility/lessons/Restore-the-omitted-digit-in-a-number-in-a-way-the-number-is-divisible-by-11.lesson>Restore the omitted digit in a number in a way that the number is divisible by 11</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/divisibility/lessons/Can-there-be-a-perfect-square.lesson>Can there be a perfect square ?</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/divisibility/lessons/Math-circle-level-problems-on-divisibility-numbers.lesson>Math circle level problems on divisibility numbers</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/divisibility/lessons/Math-circle-level-problem-on-restoring-digit-in-the-product-of-two-16-digit-numbers.lesson>Math circle level problem on restoring digit in the product of two 16-digit numbers</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/divisibility/lessons/Math-circle-level-problem-on-finding-remainders.lesson>Math circle level problem on finding remainders</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/divisibility/lessons/OVERVIEW-of-Divisibility-rules-by-2-3-4-5-6-9-10-11.lesson>OVERVIEW of Divisibility rules by 2, 3, 4, 5, 6, 9, 10 and 11</A>

Lessons that are closely adjacent to these are

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/divisibility/lessons/Product-of-two-consecutive-integers-is-divisible-by-2.lesson>Product of two consecutive integers is divisible by 2</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/divisibility/lessons/Product-of-three-consecutive-integers-is-divisible-by-6.lesson>Product of three consecutive integers is divisible by 6</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Problems-with-consecutive-odd-even-integers/Problems-dealing-with-the-product-of-two-consecutive-integers.lesson>Problems dealing with the product of two consecutive integers</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Problems-with-consecutive-odd-even-integers/Problems-dealing-with-the-product-of-three-consecutive-integers.lesson>Problems dealing with the product of three consecutive integers</A>