When learning about multiples and divisors, there are several rules of divisibility that a student may encounter. Below, we list some famous rules of divisibility:

**2:** The last digit is even.

**3:** The sum of digits is a multiple of 3.

**4:** The last 2 numbers are a multiple of 4.

**5:** The last digit is either 0 or 5.

**6:** Multiple of 2 and 3

**8:** The last 3 digits are a multiple of 8.

**9:** The sum of digits is a multiple of 9.

**10:** The last digit is 0.

**11:** The alternating sum of digits is a multiple of 11.

We will show the proofs of rules of 8 and 11. The rest follow in a similar manner.

## Proof: When a number is divisible by 8, the last 3 digits are a multiple of 8.

Let the number be \( N= 1000M + 100a + 10b + c\), where \( a, b, c\) are digits and \( M\) is a non-negative integer.

Clearly, \( 1000M\) is a multiple of 8, since \( 8 \mid 2^3 \cdot 5^3\). Hence, \( N\) is a multiple of 8 if and only if \( 100a + 10b + c\) is a multiple of 8.

## Proof: When a number is divisible by 11, the alternating sum of digits is a multiple of 11.

From Factorization , we know that

\( 10^n - (-1)^n = (10 - (-1) ) ( 10^{n-1} + 10^{n-2}(-1) + \ldots + (-1)^{n-1} )\)

is always a multiple of 11.

Hence, if \( N = 10^k a_k + 10^{k-1} a_{k-1} + \ldots + 10 a_1 + a_0\), then

\( \begin{aligned} N & = [ (10^k - (-1)^k) a_k + ( 10^{k-1} - (-1)^{k-1}) a_{k-1} + \ldots + (10+1)a_1 + (1-1)a_0 ]\\ & \quad + (-1)^k a_k + (-1)^{k-1}a_{k-1} + \ldots + (-1)^1 a_1 + (-1)^0 a_0 \end{aligned} \)

Since the terms in the square brackets consist of multiples of 11, it follows that \( N\) is a multiple of 11 if and only if the alternating sum is a multiple of 11.

## Find all possible values of \( a\) such that the number \( \overline{98a6}\) is a multiple of 3.

From the rules of divisibility, the number \( \overline{98a6}\) is a multiple of 3 if and only if the sum of the digits \( 9 + 8 + a + 6 = 23+a\) is a multiple of 3. Since \( 0 \leq a \leq 9\), this implies that \( a = 1, 4, 7\) are all the possible values.

## Show that if the last 3 digits of a number \( N\) are \( \overline{abc}\), then \( N\) is a multiple of 8 if and only if \( 4a + 2b + c\) is a multiple of 8.

This follows because \( 100a + 10 b + c = 8 (12a + b) + 4a + 2b + c\). Hence, by the divisibility rule of 8, \( N\) is a multiple of 8 if and only if \( \overline{abc} \) is a multiple of 8 if and only if \( 4a+2b+c\) is a multiple of 8.

## Divisibility rule of 7: Break up the number into blocks of 6, starting from the right. Add up all these blocks, and the resultant number has to be a multiple of 7.

This follows because \( 10^6 -1 = 999999 = 142857 \times 7\), so \( 10^{6k}M_k + 10^{6(k-1)}M_{k-1} + \ldots + 10^6 M_1 + M_0\) is a multiple of 7 if and only if \( M_k + M_{k-1} + \ldots + M_0\) is a multiple of 7. If we use \( 0 \leq M_i \leq 999999\), we get the result as stated.

## Show that the 6 digit number \( \overline{abcdef}\) is a multiple of 7 if and only if \( 5a + 4b + 6c + 2d + 3b + f\) is a multiple of 7.

Solution: This follows because \( \begin{aligned} &1 & = &0 \times 7 & + 1\\ &10 & = &1 \times 7 & + 3\\\ & 100 & =& 14 \times 7 & + 2\\ &1000 & = &142 \times 7 &+ 6\\ &10000 & = &1428 \times 7 &+ 4 \\ &100000& = &14285 \times 7 &+ 5.\\ \end{aligned} \)

Hence \( \overline{abcdef}\) is a multiple of 7 if and only if \( 5a + 4b + 6c + 2s + 3e + f\) is a multiple of 7.

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## Comments

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TopNewestDivisibility by 3: given a number in base 10 can be rewritten as:

\((a_n10^n)+(a_{n-1}10^{n-1})+(a_{n-2}10^{n-2})+...+a_1\)

We know that \(10\equiv{1(\mod{3})}\) So if we want to see if the above integer is divisible by 3, we can rewrite the integer reduced by modulo 3:

\((a_n10^n)+(a_{n-1}10^{n-1})+(a_{n-2}10^{n-2})+...+a_1\equiv{((a_n)+(a_{n-1})+(a_{n-2})+...+a_1)(\mod{3})}\)

So we will have to find the sum of the digits. QED

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For eight, I just like to use the fact that a number \(\overline {...abc} \) is a multiple of eight if

(1) \(a\) is odd, and \(\overline {bc} \) is a multiple of four, but NOT eight, (\(\overline{bc} \equiv 4 \pmod{8}\)

OR

(2) when \(a\) is even, and \(\overline {bc}\) is a multiple of eight.

This can be proved relatively simply using mod math. I find it very useful, because you need only to remember 2-digit multiples of 8 instead of 3-digit ones!

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No!not at all.You have to only look at the last 3 digits. I don't know how did you found that you have to only look at the last 2 digits.

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I'm sorry but did you even read what I wrote 0_0

My method tells you if the last three digits are a multiple of eight, but without having to memorize all three digit multiples of 8. If you read it more carefully, it should make sense, because it indeed does make sense.

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Here is a proof I wrote.

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Also, I would like to add that if a number is divisible by 11 then the alternating sum of its digits is itself divisible by 11. E.g 92818..... 9-2+8-1+8 = 22 and 11|22. Also, for 7 you take off the last digit of a number and subtract twice that number from the original. If the result is divisible by 7 then the original number is also divisible by 7 E.g. 259... 25 - 2(9) = 7 & 7|7

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Thanks for this

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