Proof of Divisibility By 6

After discussing divisibility by 5, we proceed to divisibility by 6. A number is divisible by 6 if

(1) it is even
(2) it is divisible by 3

The explanation to this is quite simple. First, if a number is even, then it is divisible by 2.

Let n be that number. Since $n$ is even, then we can write it in a form two times another integer, say $k$. That is,

$n = 2k$.

Now, looking at condition 2, a number is divisible by 6 if it is also divisible by 3. This means that $k$ must be divisible by $3$ since $2k$ is divisible by 3. Now, if $k$ is divisible by $3$, then we can write it as 3 multiplied by another integers, say, $h$. That is,

$k = 3h$.

Combining the first and second equations, we have

$n = 2k = 3(2h)$

This means that $n = 6h$ which is clearly divisible by 6.

19. April 2014 by Guillermo Bautista

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
Categories: Algebra | Tags: , , , | Leave a comment

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