# The Proof of Divisibility by 4

We examine more about divisibility rules and see why they work. In this post, we discuss divisibility by 4. If you can recall, divisibility by 4 states that

A number is divisible by 4 if the last two digits are divisible by 4.

The last two digits in that rule means the tens and ones digits.

Now, why do we only should consider the last two digits? Why is it that 76200348 divisible by 4 just because 48 is divisible by 4.

If we examine the structure of numbers, every whole number can be broken down into the representation $a(100) + b(10) + c$

For example, 325 can be represented $3(100) + 2(10) + 5$ and $46781$ can be represented as $467(100) + 8(10) + 1$

Now, here is the trick. Since 100 is divisible by 4, then a(100) part is always divisible by 4. That is why, we only have to check the last two digits.

Algebraically, if $10b + c$ is divisible by $4$, then $\displaystyle \frac{10b + c}{4}$

has an integer quotient $k$ which means that $\frac{10b + c}{4} = k$

Multiplying both sides by $4$ we get $10b + c = 4k$.

This means that we can replace $10b + c$ in the original representation by $4k$. This makes $100a + 10b + c = 100a + 4k$.

Notice that in the right hand side of the equation, $100a + 4k$, we can factor out $4$. That is, $100a + 4k = 4(25a + k)$.

Now, the representation $4(24a + k$means that the given above number is divisible by 4. This confirms the rule that you just have to check the last two digit number of any number to see if it is divisible by 4.