## The Proof of Divisibility By 5

This is the fourth post in the divisibility series for middle school students. In this post, I am going to discuss divisibility by 5. This is probably one of the easiest divisibility rules to remember and to prove. I am sure you are all familiar that a number ends in 0 or 5, then it is divisible by 5. Therefore, we have two cases.

Theorem

If a number ends in 0 or 5, then it is divisible by 5.

Proof

Numbers Ending in 0

All numbers ending in 0 can be expressed as 10n where n is some numbers. For example, 30 can be expressed as 10(3) and 120 can be expressed as 10(12). Now, since 10 is divisible by 5, so 10n is divisible by 5. This means that all numbers ending in 0 are divisible by 5.  Continue Reading →

09. April 2014 by Guillermo Bautista

## 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. Continue Reading →

08. March 2014 by Guillermo Bautista

## The Proof of the Divisibility by 3

In the previous post, we have discussed about divisibility by 2. In this post, we discuss about divisibility by 3.

Rule: A number is divisible by 3 if the sum of the digits is divisible by 3.

The number 321 is divisible by 3 because 3 + 2 + 1 = 6 is divisible by 3. On the other hand, the number 185 is not divisible by 3 because 1 + 8 + 5 = 14 is not divisible by 3. Now, why does this rule work?

Notice how the numbers are represented in expanded notation:

$534 = 5(100) + 3(10) + 4$

$185 = 1(100) +8(10) + 5$

This means that number in hundreds can be represented as

$100h + 10t + u$

where h, t, u are the hundreds, tens, and units digits. Now, we can represent $100h + 10t + u$ as

$(99h + h) + (9t + t) + u$ and regroup the terms as

$(99h + 9t) + (h + t + u)$.

Of course, $99h + 9t$ is divisible by 3, so it only remains to show that $h + t + u$ is divisible by 3. But, $h + t + u$ is the sum of the digits of a 3-digit number. This proves (for three digit numbers) that the rule above is true.

Although the proof above works only for 3-digit numbers,  it can be done to any number of digits.

27. February 2014 by Guillermo Bautista