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# Divisibility Rules

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:

### When dividing by...

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.

## Application and Extentions

### 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.

Note by Arron Kau
3 years, 6 months ago

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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 · 1 year, 7 months ago

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! · 3 years, 1 month ago

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. · 3 years, 1 month ago

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. · 3 years, 1 month ago

I don't think so. · 3 years, 1 month ago

Here is a proof I wrote. · 3 years, 1 month ago

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 · 2 years, 10 months ago