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A University-Level Understanding of Number with an eye towards elementary education

Shannon Ezzat

Section 7.2 Divisibility Theorems for Small Numbers

Subsection 7.2.1 Introduction

In this section, we will introduce and prove the divisibility theorems for numbers 2 through 11. Each theorem will be introduced with an example, and then we will prove them for four-digit numbers. While we will focus on four-digit numbers for simplicity, it’s important to note that these proofs can be extended to numbers of any number of digits.
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To facilitate our discussion, we will use underline notation in LaTeX to indicate a number by its place value digits. For example, a four-digit number \(ABCD\) can be represented as \(\underline{ABCD} = A \times 1000 + B \times 100 + C \times 10 + D\text{.}\) This notation will allow us to break down the number into its constituent parts and analyze its divisibility properties. Since we are using variables to represent unknown digits, we need this new notation, since if we write \(ABCD\) this usually means \(A \times B \times C \times D\text{.}\) However, if we write actual digits like \(1234\) without an underline, we still mean \(1000+200+30+4\text{.}\)
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In essence, the idea of divisibility rules is to turn a question about divisibility of a large number into a question about divisibility of a smaller number. In some cases, that number will have a fixed number of digits, and in other cases the number of digits won’t be fixed, but it will be a smaller number.
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Subsection 7.2.2 Divisibility Rules for 2, 4, and 8

In number theory, divisibility rules are simple methods to determine whether one number is divisible by another without performing the division. This section will focus on the divisibility rules for 2, 4, and 8.
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Subsubsection 7.2.2.1 Divisibility Rule for 2

Most people reading this will know the divisibility rule for \(2\text{.}\) It is one of the first patterns that elementary students notice, and of course it extends to being able to check if any number has 2 as a factor. The rule is as follows; we will prove it after showing examples of the rule showing both divisibility and non-divisibility.
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Reminder.
A reminder from the first chapter. Remember that an if-and-only-if statement, say \(P\) if-and-only-if \(Q\) is shorthand to mean "\(P \to Q\) and \(Q \to P\)". In fact, since the contrapositive of \(Q \to P\) is \(\sim P \to \sim Q\text{,}\) we can view \(P\) if-and-only-if \(Q\) as "\(P \to Q\) and \(\sim P \to \sim Q\)".
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A number is divisible by 2 if and only if the last digit is divisible by 2.
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Example 7.2.1. Example.
Consider the number 1234. The last digit is 4, which is even. Therefore, 1234 is divisible by 2.
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Example 7.2.2. Non-Example.
Consider the number 1235. The last digit is 5, which is not even. Therefore, 1235 is not divisible by 2.
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Think about why we would only need to check the last digit of our number. It may be helpful to write \(1234\) as \(1000+200+30+4\) and \(1235\) as \(1000+200+30+5\text{.}\) Could you explain why this rule is true to another student?
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Let’s now prove this rule for four-digit numbers, using the notation mentioned in the introduction. We will rely on the divisibility theorems in the previous section to justify why this rule is true.
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Proof.
Consider the four-digit number \(\underline{ABCD}=A\times 1000+B\times 100+C\times 10+D\text{.}\) We know that \(2 \mid 1000\text{,}\) \(2 \mid 100\text{,}\) and \(2 \mid 10\text{,}\) so by Subsubsection 7.1.2.1 we know that \(2 \mid A\times 1000,\) \(2 \mid B\times 100\text{,}\) and \(2 \mid C \times 10\text{.}\) By Subsubsection 7.1.2.2 we know that \(2 \mid (A\times 1000+B\times 100+C\times 10).\\text{.}\) For ease, let’s define \(S:=A\times 1000+B\times 100+C\times 10\text{.}\) Thus we can write \(\underline{ABCD}=S+D\text{.}\)
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Now with that setup we can start proving \(P \to Q\) and \(\sim P \to \sim Q\text{.}\) By Subsubsection 7.1.2.2 we know that if \(2 \mid D\) (P being true) then \(2 \mid S+D\text{,}\) so Q is true. So we have that \(P \to Q\) is always true. By Subsubsection 7.1.2.3 if \(2 \nmid D\) (P being false) then \(2 \nmid S+D\text{,}\) so Q is false. So we have that \(\sim P \to \sim Q\) is always true. INSERT PIC OF THIS PROOF
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Subsubsection 7.2.2.2 Divisibility Rule for 4

The divisibility rule for 4 is very similar to the divisibility rule for 2. For the previous rule, we knew that \(2\) was a factor of every power of ten \(10^e\) with \(e \geq 1\text{.}\) It is a good idea to see if you can modify the work from the divisibility rule for 2 to develop a divisibility rule for 4. Try this yourself before reading on.
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Again, we state the rule and give some examples and non-examples before we prove that the rule always works.
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A number is divisible by 4 if and only if the number formed by the last two digits is divisible by 4.
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Example 7.2.4. Example.
Consider the number 1232. The number formed by the last two digits is 32. Since 32 is divisible by 4, 1232 is also divisible by 4.
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Example 7.2.5. Non-Example.
Consider the number 1235. The number formed by the last two digits is 35. Since 35 is not divisible by 4, 1235 is not divisible by 4.
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Proof.
Consider the four-digit number \(\underline{ABCD}=A\times 1000+B\times 100+C\times 10+D\text{.}\) We know that \(4 \mid 1000\) and \(4 \mid 100\text{,}\) so by Subsubsection 7.1.2.1 we know that \(4 \mid A\times 1000,\) and \(4 \mid B\times 100\text{.}\) By Subsubsection 7.1.2.2 we know that \(4 \mid (A\times 1000+B\times 100).\\text{.}\) For ease, let’s define \(S:=A\times 1000+B\times 100\text{.}\) Thus we can write \(\underline{ABCD}=S+(C\times 10+D)\text{.}\)
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Now with that setup we can start proving \(P \to Q\) and \(\sim P \to \sim Q\text{.}\) By Subsubsection 7.1.2.2 we know that if \(4 \mid (C\times 10+D)\) (P being true) then \(4 \mid S+C\times 10+D\text{,}\) so Q is true. So we have that \(P \to Q\) is always true. By Subsubsection 7.1.2.3 if \(4 \nmid (C\times 10+D)\) (P being false) then \(4 \nmid S+(C\times 10+D)\text{,}\) so Q is false. So we have that \(\sim P \to \sim Q\) is always true. INSERT PIC OF THIS PROOF
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Subsubsection 7.2.2.3 Divisibility Rule for 8

Looking at the above rules, you may have started to notice a pattern emerging. We can view \(2\) as \(2^1\) and note that to check divisibility by 2 we only need to check whether the number formed by the last digit is divisible by 2. For \(4=2^2\text{,}\) we need to check if the number formed by the last *two* digits is divisible by \(4\text{.}\) Let’s check if the pattern has a chance of holding for \(8=2^3:\)
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Example 7.2.7.
Consider the number 1232. The number formed by the last three digits is 232. We can check that \(232 \div 8 = 29\text{,}\) and \(1232 \div 8 = 154\text{.}\) So it looks like the expected rule holds, at least for this example. Indeed, since 232 is divisible by 8, 1232 is also divisible by 8.
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Here’s a non-example too:
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Example 7.2.8. Non-Example.
Consider the number 1235. The number formed by the last three digits is 235. Since 235 is not divisible by 8, 1235 is not divisible by 8.
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Proof.
Consider the four-digit number \(\underline{ABCD}=A\times 1000+B\times 100+C\times 10+D\text{.}\) We know that \(8 \mid 1000\text{,}\) so by Subsubsection 7.1.2.1 we know that \(8 \mid A\times 1000\text{.}\) For ease, let’s define \(S:=A\times 1000\text{.}\) Thus we can write \(\underline{ABCD}=S+B\times 100+C\times 10+D\text{.}\)
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Now with that setup we can start proving \(P \to Q\) and \(\sim P \to \sim Q\text{.}\) By Subsubsection 7.1.2.2 we know that if \(8 \mid B\times 100+C\times 10+D\) (P being true) then \(8 \mid S+B\times 100+C\times 10+D\text{,}\) so Q is true. So we have that \(P \to Q\) is always true. By Subsubsection 7.1.2.3 if \(8 \nmid B\times 100+C\times 10+D\) (P being false) then \(8 \nmid S+B\times 100+C\times 10+D\text{,}\) so Q is false. So we have that \(\sim P \to \sim Q\) is always true. INSERT PIC OF THIS PROOF
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Example 7.2.10. Example.
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Subsubsection 7.2.2.4 Divisibility Rule for Higher Powers of 2

You can see that there seems to be a pattern emerging here. A good question is to see if we can figure out why this pattern holds. Let’s walk through this in an exercise:
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Checkpoint 7.2.11.
  1. What is the largest power of \(2\) that is a factor of \(10^3\text{?}\) What about \(10^7\text{?}\) What about \(10^k\) for some number \(k\text{?}\) Try factoring \(10^k\) into its prime factors.
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  2. We know that \(8 \mid 1000\text{.}\) What can you say about whether \(8 \mid 10^k\) for \(k \geq 4\text{?}\) What theorem that we’ve proved justifies this?
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  3. .
    Can you generalize the above idea to all powers of \(2\text{?}\) What can you say about which powers of \(10\) it is a factor of?
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  4. How does this relate to digits of a number?
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  5. Using the ideas above, conjecture a divisibility rule for when a number is divisible by \(2^k\text{.}\) What do you have to check?
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  6. Let’s say you have an \(n\)-digit number \(\underline{A_{10^{n-1}}A_{10^{n-2}}\ldots A_{10}A_1}\) where the subscript of each digit is its place value. Using the proofs for \(2,4,\) and \(8\) as a blueprint, prove the conjecture you made above.
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Subsection 7.2.3 Divisibility Rule For 5

The divisibility rule for 5 is also incredibly well known. Most people reading this should know that you only need to check whether the last digit of the number is either 5 or 0. However, like the divisibility rule for 2, it will be more advantageous to think the rule as the last digit being divisible by \(5\text{.}\) Note that the rule for divisibility by 2 relies on the fact that \(10=2\times5\text{,}\) since all multiples of \(10\text{,}\) including powers of \(10\text{,}\) are also divisible by \(2\text{.}\) Since \(10=2 \times 5,\) the exact same idea should work for \(5\) as well as \(2\text{.}\) In fact, we should be able to write down the exact same argument for divisibility by 5 as divisibility by 2, with all the 2’s replaced by 5:
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Proof.

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Example 7.2.13.

Consider the number \(3456\text{.}\) Since \(5 \nmid 6\text{,}\) we know that \(5 \nmid 3456\text{.}\) You can check this using long division or a calculator.
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Example 7.2.14.

Consider the number \(2345\text{.}\) Since \(5 \mid 5\text{,}\) we know that \(5 \mid 2345\text{.}\) Again, you can check this via long division.
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Checkpoint 7.2.15.

Try and modify the divisibility rules for powers of \(2\) to be divisibility rules for powers of \(5\text{.}\) What do you have to check to determine if a number is divisible by \(5^2=25\text{?}\) What about \(5^3=125\text{?}\) What about \(5^k\) for some number \(k\text{?}\)
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Subsection 7.2.4 Divisibility Rules For 3 And 9

In the last section, we were able to develop divisibility rules for powers of \(2\) and \(5\) since powers of \(10\) are divisible by \(2\) and \(5\text{.}\) We could use this commonality between each power of \(10\text{,}\) and thus each place value in a number, to come up with a simplified rule for determining divisibility of these powers of \(2\) and \(5\text{.}\) In this section, we will look for another consistent commonality between powers of 10 being divided by \(3\) and \(9\) that will allow us to develop divisibility rules for these numbers. The next example will walk you through noticing the pattern.
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Example 7.2.16.

In this example we will develop an intuition for a divisibility rule for \(3\) and \(9\text{.}\)
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  1. Using long division, so you can keep track of remainders, find \(10 \div 3\text{.}\) Now do the same thing for \(100, 1000, 10000,\) and \(100000\text{.}\)
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  2. What commonality do you notice about these answers? In each division, how many of each power of 10 are you able to group into groups of 3?
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  3. Now do the division \(111 \div 3\) in two ways; using long division, and then doing the division separately for each place value; that is, \((100+10+1) \div 3 = 100\div 3 + 10 \div 3 + 1\div 3.\) What can you say about the three remainders in the second case?
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  4. Using the above as a guide, describe a way to determine if a number is divisible by \(3\text{.}\)
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  5. Now redo the work above replacing \(3\) by \(9\text{.}\) You should notice the exact same pattern.
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Now that we have developed our rule let’s prove it for four-digit numbers. Like the other proofs in this section, this can be expanded to a proof for number with any amount of digits.
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Proof.

Consider a number \(\underline{abcd}\) where \(a\text{,}\) \(b\text{,}\) \(c\text{,}\) and \(d\) are its digits. We can express this number in its expanded form as:
\begin{equation*} \underline{abcd} = a \times 1000 + b \times 100 + c \times 10 + d \end{equation*}
Using place value, we can rewrite this as:
\begin{equation*} \underline{abcd} = a \times (999 + 1) + b \times (99 + 1) + c \times (9 + 1) + d \end{equation*}
Expanding and grouping, we get:
\begin{equation*} \underline{abcd} = a \times 999 + b \times 99 + c \times 9 + (a + b + c + d) \end{equation*}
The terms \(a \times 999\text{,}\) \(b \times 99\text{,}\) and \(c \times 9\) are all divisible by \(3\) and \(9\) since \(999\text{,}\) \(99\text{,}\) and \(9\) are divisible by both \(3\) and \(9\text{.}\) Therefore, the divisibility of the number \(\underline{abcd}\) by \(3\) or \(9\) depends solely on the sum of its digits, \(a + b + c + d\text{.}\)
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Let’s look at a few small examples to make sure we know how the rules work.
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Example 7.2.18. Example for Divisibility by 3.

Consider the number 123. The sum of its digits is \(1 + 2 + 3 = 6\text{.}\) Since 6 is divisible by 3, the number 123 is also divisible by 3.
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Example 7.2.19. Example for Divisibility by 9.

Consider the number 234. The sum of its digits is \(2 + 3 + 4 = 9\text{.}\) Since 9 is divisible by 9, the number 234 is also divisible by 9.
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Example 7.2.20. Non-example for Divisibility by 3.

Consider the number 124. The sum of its digits is \(1 + 2 + 4 = 7\text{.}\) Since 7 is not divisible by 3, the number 124 is not divisible by 3.
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Example 7.2.21. Non-example for Divisibility by 9.

Consider the number 235. The sum of its digits is \(2 + 3 + 5 = 10\text{.}\) Since 10 is not divisible by 9, the number 235 is not divisible by 9.
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With smaller examples like the above, it might be difficult to see why these rules are very useful and powerful. Let’s look at another, more difficult example:
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Example 7.2.22.

A student wrote down the following ten digit number: \(1,435,3k7,451\text{,}\) where \(k\) is some digit that was too messy in their notes to make it out. However, the student does know that the number is divisible by 3 but not divisible by 9. What are the possible choices for the digit \(k\text{?}\)
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Solution.
This would be a very hard problem using long division, and even using a calculator naively would require ten different trials, one for each digit. However, we can use the divisibility rules above to help us out.
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We know that the number is divisible by 3, so the sum of the digits must also be divisible by 3. The sum of the digits is \(1+4+3+5+2+k+7+4+5+1=32+k\text{.}\) We know that we can add either \(1,4,\) or \(7\) to \(32\) to get \(33,36,\) or \(39\text{,}\) which are multiples of \(3\text{.}\) But since the number isn’t divisible by 9, we know that the sum cannot be divisible by 9, so it cannot be \(36\text{.}\) So \(k \neq 4\text{.}\) So the digit is either \(1\) or \(7\text{.}\)
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Subsection 7.2.5 Divisibility Rule For 11

There is a nice divisibility rule for 11 that is similar to the divisibility rules for 3 and 9. The key of the divisibility rule for 3 and 9 was that for any \(k \in \mathbb{N}_0\text{,}\) it’s true that \(10^k \div 3\) has a remainder of \(1\text{.}\) Let’s explore dividing powers of \(10\) by \(11\) to see if we notice any patterns.
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Checkpoint 7.2.23.

  1. Let’s start with \(10^0=1\text{.}\) What is the remainder of \(10^0 \div 11\text{?}\) It will be helpful to keep this information in a table.
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  2. Now do the same thing for \(10^1, 10^2, 10^3\) and \(10^4\text{.}\) That is, calculate the remainder of \(10^k \div 11\) for \(k=1,2,3,4\text{.}\)
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  3. Make a conjecture about the patter of remainders of \(10^k \div 11\) for each \(k \in \mathbb{N}_0\text{.}\)
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  4. Using the area and grouping model of division, draw a diagram that shows a few examples of this conjecture. Can you express the "large" remainders in terms of what is missing to obtain a full group of 11?
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Hopefully the exercise above gives you some intuition about what a divisibility rule for 11 will require. Note that there powers of 10, and therefore place values, fall into two groups; a remainder of 1 when \(k\) is even, and a remainder of 10 (or a "remainder" of -1) when \(k\) is odd. So when \(k\) is even, \(11 \mid (10^k-1)\text{,}\) and when \(k\) is odd \(11 \mid (10^k+1)\text{.}\) So we can express powers of 10 as numbers "close to" numbers divisible by 11, plus or minus 1. This is the key for this rule. Let’s prove this now for six-digit numbers, just so we get a feel for the pattern that is happening.
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Proof.

Let’s prove this for a six-digit number, but the proof can be easily extended to a number of any number of digits. Consider a six-digit number \(\underline{ABCDEF}\) where \(A\text{,}\) \(B\text{,}\) \(C\text{,}\) \(D\text{,}\) \(E\text{,}\) and \(F\) are its digits. We can express this number in its expanded form as:
\begin{equation*} \underline{ABCDEF} = A \times 100000 + B \times 10000 + C \times 1000 + D \times 100 + E \times 10 + F \end{equation*}
Using properties of powers of 10 and their relation to 11, we can rewrite this as:
\begin{equation*} \underline{ABCDEF} = A \times (10^5 + 1) - A + B \times (10^4 - 1) + B + C \times (10^3 + 1) - C + D \times (10^2 - 1) + D + E \times 10 + F \end{equation*}
Grouping the terms, we get:
\begin{equation*} \underline{ABCDEF} = A \times 10^5 + B \times 10^4 + C \times 10^3 + D \times 10^2 + E \times 10 + F - (A - B + C - D + E - F) \end{equation*}
The terms \(A \times 10^5\text{,}\) \(B \times 10^4\text{,}\) \(C \times 10^3\text{,}\) \(D \times 10^2\text{,}\) and \(E \times 10\) are all divisible by 11 due to the properties of powers of 10. Therefore, the divisibility of the number \(\underline{ABCDEF}\) by 11 depends solely on the alternating sum of its digits, \(A - B + C - D + E - F\text{.}\)
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Subsection 7.2.6 Compound Divisibility Rules

Subsubsection 7.2.6.1 When Compound Divisibility Rules Work

Let’s say you are interested in divisibility by some number \(k\text{,}\) and \(k=a \times b\text{.}\) This means that \(a \mid k\) and \(b \mid k\text{.}\) So if for some number \(n\) we know \(k \mid n\text{,}\) then by Theorem 7.1.4 we know that \(a \mid n\) and \(b \mid n\text{.}\) So we can say if \(k \mid n\) then \(a \mid n\) and \(b \mid n\text{.}\)
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Does the converse of this hold as well? That is if we know \(a \mid n\) and \(b \mid n\) then do we know that \(k \mid n\text{?}\) Let’s explore with some examples and see if we can determine the conditions when we can indeed combine divisibility rules to make a new divisibility rule.
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Checkpoint 7.2.25.
  1. Let’s start to think about numbers divisible by both, noting that \(6=2 \times 3\text{.}\) Using a calculator if you wish, write down around 10 or 20 examples of numbers with both 2 and 3 as factors (can you think about how to easily make these?). Are these numbers also divisible by 6? Check!
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  2. If you haven’t found one above, can you think of a number that is divisible by 2 and 3 but not divisible by 6?
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  1. Let’s try another number, say \(24=6 \times 4\text{.}\) Write down many numbers with both 6 and 4 as factors. Is 24 also a factor of these numbers? Check!
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  2. If you haven’t found one above, can you think of a number that is divisible by 6 and 4 but not divisible by 24?
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  3. Try the above with other sets of numbers \(a\) and \(b\text{,}\) for example try \(3 \times 5=15\) and \(2 \times 4=8\text{.}\) Do you notice a pattern about the numbers \(a\) and \(b\) where \(a \times b\) is always a factor of numbers where both \(a\) and \(b\) are factors, and when there are numbers that do not have \(a \times b\) as a factor?
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In the above example you probably have seen that if \(a\) and \(b\) have no common factors, then \(a \times b\) is necessarily a factor of any number where both \(a\) and \(b\) are factors. Let’s try and prove this.
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Proof.
Since this is a biconditional statement, we need to show that \(P \to Q\) and \(Q \to P\text{.}\) The direction \(P \to Q\) was already proved in the first paragraph of this subsection. So let’s do the other direction.
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Let’s assume the "if-part" is true: let \(a \mid n\) and \(b \mid n\text{.}\) We need to show that the "then-part" is true, that is, \(ab \mid n\text{.}\)
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Since we know \(a \mid n\) we know that \(n=ar\) for some natural number \(r\text{.}\) But we also know \(b \mid n\text{,}\) so since \(n=ar\) we know \(b \mid ar\text{.}\) Since \(ar\) has a factor of \(b\text{,}\) and \(a\) and \(b\) share no common factors, then all of the factors of \(b\) must be in \(r\text{,}\) so \(r=bs\) for some number \(s\text{.}\) So \(n=ar = a(bs) = (ab)s\text{.}\) So since we’ve written \(n\) as a multiple of \(ab\text{,}\) this means that \(ab \mid n\text{.}\)
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To show that combining divisibility rules for numbers \(a\) and \(b\) doesn’t work if \(a\) and \(b\) have a common factor, let’s try and develop a counterexample for each choice of two numbers with a common factor.
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Checkpoint 7.2.27.
For the following pairs of numbers, find the smallest number \(n\) where \(a \mid n\) and \(b \mid n\) but \(ab \nmid n\text{.}\) What is the largest common factor of \(a\) and \(b\text{.}\)
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  1. \(a=2\) and \(b=4\text{.}\)
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  2. \(a=6\) and \(b=4\text{.}\)
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  3. \(a=10\) and \(b=15\)
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  4. Do you notice a way of determining the smallest counterexample using the numbers \(a\) and \(b\) and their greatest common factor?
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  5. Try and prove that your method above works in general.
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Subsubsection 7.2.6.2 Some Common Compound Divisibility Rules

From the subsection above, we know that as long as two numbers share no common factors, we can indeed combine their divisibility rules to determine if a number is divisible by their product. Let’s list a few of these for small numbers, and give an example and non example for each.
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  1. Rule for Divisibility by 6: A number is divisible by 6 if it is divisible by both 2 and 3.
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    Example 7.2.28.
    12 is divisible by 6 because it is even (divisible by 2) and the sum of its digits (1 + 2 = 3) is divisible by 3.
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    Example 7.2.29.
    14 is not divisible by 6 because while it is even (divisible by 2), the sum of its digits (1 + 4 = 5) is not divisible by 3.
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  2. Rule for Divisibility by 10: A number is divisible by 10 if its last digit is 0; that is, it is divisible by 2 and divisible by 5.
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    Example 7.2.30.
    150 is divisible by 10 because its last digit is 0.
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    Example 7.2.31.
    153 is not divisible by 10 because its last digit is 3, not 0.
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  3. Rule for Divisibility by 15: A number is divisible by 15 if it is divisible by both 3 and 5.
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    Example 7.2.32.
    30 is divisible by 15 because the sum of its digits (3 + 0 = 3) is divisible by 3 and its last digit is 0 or 5 (indicating divisibility by 5).
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    Example 7.2.33.
    35 is not divisible by 15 because while its last digit is 5 (indicating divisibility by 5), the sum of its digits (3 + 5 = 8) is not divisible by 3.
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Subsection 7.2.7 Iterative Divisibility Rules

Sometimes, to determine if a number is divisible by another, we can use an iterative approach. This means applying the divisibility rule repeatedly until we get a number that we can easily determine its divisibility. This method is particularly useful for large numbers and for certain divisibility rules like those for 3, 9, and 11, which turn the problem of checking divisibility of a large number into checking divisibility of a smaller number.
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Let’s explore this iterative method with examples and non-examples for divisibility by 3, 9, and 11.
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Subsubsection 7.2.7.1 Divisibility by 3

Example 7.2.34.
To check if \(3\) divides \(12345678901\text{,}\) we sum its digits: \(1+2+3+4+5+6+7+8+9+0+1 = 46\text{.}\) Now, to check if \(3\) divides \(46\text{,}\) we sum its digits: \(4+6 = 10\text{.}\) Since 10 is not divisible by 3, neither is \(12345678901\text{.}\)
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For \(9876543210\text{,}\) the sum of its digits is \(45\text{,}\) which is divisible by 3. Therefore \(9876543210\) is indeed divisible by 3.
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Subsubsection 7.2.7.2 Divisibility by 9

Example 7.2.35.
To check if \(9\) divides \(123456789012\text{,}\) we sum its digits: \(1+2+3+4+5+6+7+8+9+0+1+2 = 48\text{.}\) Now, to check if \(9\) divides \(48\text{,}\) we sum its digits: \(4+8 = 12\text{.}\) Finally, summing the digits of 12 gives \(3\text{,}\) which is not divisible by 9. Thus, \(123456789012\) is not divisible by 9.
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For \(98765432109\text{,}\) the sum of its digits is \(54\text{,}\) which is divisible by 9. Therefore \(98765432109\) is indeed divisible by 9.
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Subsubsection 7.2.7.3 Divisibility by 11

Example 7.2.36.
To check if \(11\) divides \(1234567890123\text{,}\) we alternate the sum and subtraction of its digits: \(1-2+3-4+5-6+7-8+9-0+1-2+3 = 17\text{.}\) Since 17 is not divisible by 11, neither is \(1234567890123\text{.}\)
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For \(9090909090909090909090909040\text{,}\) the alternating sum is \(121\text{,}\) and the alternating sum of \(121\) is \(0\text{.}\) Therefore \(9090909090909090909090909040\) is divisible by 11.
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Subsection 7.2.8 MakingDivisibilityRules

MAKING A DIVISIBILITY RULE FOR ANY NUMBER USING TECHNIQUE FROM WIKIPEDIA.
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