# Divisibility Tests

These can be derived by taking successive powers of 10 modulo the number you wish to create a rule for.

$10^0 \equiv 1 \pmod{7}$
$10^1 \equiv 3 \pmod{7}$
$10^2 \equiv 2 \pmod{7}$
$10^3 \equiv 6 \pmod{7}$
$10^4 \equiv 4 \pmod{7}$
$10^5 \equiv 5 \pmod{7}$
$10^6 \equiv 3 \pmod{7} \cdots$
A full explanation will have to wait for another day.

Ways of telling whether an integer is divisible by:

### 1

Drop an apple. If it falls downwards, the integer is divisible by 1.

### 2

Yes, if the last digit is divisible by two.

### 3

Yes, if the sum of the digits is divisible by 3.

### 4

Yes, if the last digit of the integer divided by two is divisible by two.

### 5

Yes, if the last digit is a 0 or a 5.

### 6

Divide the integer by 2. Yes, if the sum of the digits of the result is divisible by three.

### 7

Let the digits of the number be $a_0, a_1, a_2, a_3, \cdots a_k$ where $a_0$ has the lowest place value. The number is divisible by 7 if 7 divides the sum of $a_0 + 3a_1 + 2a_2 + 6a_3 + 4a_4 + 5a_5 + \cdots \\ + a_{(k)} + 3a_{(k+1)} + 2a_{(k+2)} \cdots$
Example: Does 7 divide 3236655
$a_0 + 3a_1 + 2a_2 + 6a_3 + 4a_4 + 5a_5 + a_6$
$= 3 + 3 \times 5 + 2 \times 6 + 6 \times 6 + 4 \times 3 + 5 \times 2 + 3$
$= 3 + 15 + 12 + 36 + 12 + 10 + 3 = 91$
and
$1 + 3 \times 9 = 28$ and as 7 clearly divides 28, 7 divides 91, hence 7 divides 3236655.

### 9

Yes, if the sum of the digits is divisible by 9.

## 2 thoughts on “Divisibility Tests”

1. Seb says:

6:

Divide the integer by 2. Yes, if the sum of the digits of the result is *divisible by* three.

:)

2. jebavarde says:

Thanks :D