# Divisibility by 4

This is the second post in the Divisibility Rules Series. In the last post, we discussed about divisibility by 2. In this post, we discuss divisibility by 4.

Now, how do we know if a number is divisible by $4$?

Four divides $100$ because $4 \times 25 = 100$. It is also clear that four divides $200, 300, 400$ and all multiples of $100$. Therefore, four divides multiples of $1000$, $10 000$, and $100 000$. In general, $4$ divides $10^n$, where $n$ is an integer greater than $1$.

Now, how do we know if a number that is not a power of $10$ is divisible by $4$. Let us try a few examples.

Example 1: Is $148$ divisible by $4$? $148$ is equal to $100 + 48$ and $100$ is divisible by $4$. Since $48$ is also divisible by $4$, therefore, $148$ id divisible by $4$.

Example 2: Is $362$ divisible by $4$? $362$ is equal to $300 + 62$. Now, $300$ is divisible by $4$. Since $62$ is not divisible by $4$, therefore, $362$ is not divisible by $4$.

Example 3: Is $3426$ divisible by $4$? $3426 = 3400 + 26$. Now, $3400$ is divisible by $4$ (it’s a multiple of 100), and $26$ is not divisible by $4$. Therefore, $3426$ is not divisible by $4$.

By now, you would have realized that we just test the last 2 digits of the numbers if we want to find out if it is divisible by 4: 148, 362, and 3426.

The Algebraic Explanation

We can always expand any number as product of integers and powers of $10$. For example $148$ and $3426$ maybe represented as follows: In general, any whole number $a_{n}a_{n-1}\cdots a_2a_1a_0$ with digits $a_n$, $a_{n-1}$, $\cdots$, $a_2$, $a_1$ and $a_0$ can be  expressed as and all numbers at the left hand side of $a_2(10^2)$ are divisible by $4$ since they are multiples of $10^n$ where $n$ is an integer greater than 1. Therefore, we just have to look at $a_1(10^1) + a_0(10^0)$, or the last two digits (the tens and the ones) of the numbers to see if it is divisible by $4$.

Hence, we have the following rule: A number is divisible by $4$ if the last two digits are multiples of $4$. 