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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TopNewestFor 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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Divisibility 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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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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