Prime Factorization by Upside-Down Division — Approach & Examples

Find Prime Factors by Upside-Down Division

We have a new way to find the prime factorization of a number! This is another method that can be used to find the factors of a number. In this method, you will divide a number evenly by the smallest prime number you can.

It is called the upside down division because the division symbol is flipped upside down. This method is in lieu of using a factor tree.

Let’s take a look at an example with $40$. We'll try to find the prime factors of $40$.

The first prime is $2$, and divides $40$ evenly. This can be seen with a divisibility trick. Most importantly is that $40$ is even. All even numbers are multiples of $2$.

We can write $2$ on the left side of the number and divide $40$ by $2$, writing the result below.

We continue by finding the smallest prime number that is a factor of $20$.

The smallest prime number is $2$ again, so we write $2$ on the left side again, and write $20 \div 2 = 10$ below.

We see that $20$ is a composite number, so we're not done yet. We continue this process until the number on the bottom is a prime number.

Here is our final result. Now, we take all the numbers on the outside left and the number on the bottom, and this is our prime factorization. $$40 = 2 \times 2 \times 2 \times 5$$ We also have the option to present the final factorization as a product of exponential numbers having a prime number base in the exponential notation. $$40 = 2^3 \times 5$$ We are now going to take a look at some of the types of problems you might encounter about finding prime factors by upside down division. Before you start these problems, make sure to go over the basics of finding prime factors by upside down division.

One more note: you can always check a prime factorization. Multiply all the numbers you have in your factorization. Your answer should be the number you started with.

Here is our final result. Now, we take all the numbers on the outside left and the number on the bottom, and this is our prime factorization. $$\boxed{56 = 2 \times 2 \times 2 \times 7}$$

Interactive Problem

Before you begin, make sure you know how to do upside-down prime factorization.

Here's a graphic to remind you of how to use this method:

Now, let's begin.

Wow, excellent choice right there!

So we start by writing our number down and drawing our upside down division box.

Yes! The number ends with an even number so dividing by $2$ is a good first choice.

Yes! The number ends with an even number so dividing by $2$ is a good first choice.

Yes! The number ends with an even number so dividing by $2$ is a good first choice.

Yes! If we add the digits up we get $24$ which is divisible by $3$.

Yes! $7$ is the next best prime to divide by. It might not seem apparent. If you're unsure you can always quickly divide to the side to see if it goes in.

Yes! $7$ goes into this number.

Yes, $17$ is a prime number so we stop here.

Yes! We include all the numbers to the left and the one at the bottom.

(Video) Prime Factorization

by mathman1024

This video by mathman1024 teaches prime factorization through the upside down division method.

$\large{\text{Summary}}$

He uses an example with $120$. The first prime is $2$. Does $2$ go into $120$ evenly? Yes it does! We write $2$ on the left side of the number and then divide $120$ by it, writing the result below.

We continue by finding the smallest prime that goes into $60$.

We continue this process until the bottom number is prime.

The final result is $5$. We can take all the numbers and write out our prime factorization. $$\begin{align}120&=2 \cdot 2 \cdot 2 \cdot 3 \cdot 5\\ 120&= 2^3 \cdot 3 \cdot 5 \end{align}$$

These two ways of writing $120$ are equivalent. It just so happens the second way uses exponents. Either way we've found the set of prime factors of $120$. It's just the numbers being multiplied together!