# Explain what is happening when using the difference method for determining the greatest common factor. Why does this work?

## Use a numeric reference to compare and check the presented logic.

Mar 1, 2017

See the explanation

#### Explanation:

$\textcolor{b l u e}{\text{The numeric reference}}$

Let one of the common factors be $f = 8$
let a numeric count be $n$

As the numbers to be tested I chose:

$8 \times 20 = 160$
$8 \times 15 = 120$
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$\textcolor{b l u e}{\text{The underlying principle}}$

As the process is based on subtraction then the starting point of

$160 - 120$ has to have a difference that is related to one of the factors. In that: $\text{ "120+nxx"some factor of 160} = 160$

This will be true of every subtraction in that the difference will a factor.

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$\textcolor{b l u e}{\text{The demonstration of process}}$

Set the following
$8 \times 20 = 160 = 20 f$
$8 \times 15 = 120 = 15 f$

The subtraction process

$20 f - 15 f = \textcolor{w h i t e}{1} 5 f \leftarrow \text{ largest - smallest: next use the 15 & 5}$

$15 f - \textcolor{w h i t e}{1} 5 f = 10 f \leftarrow \text{ largest - smallest: next use the 10 & 5}$

$10 f - \textcolor{w h i t e}{1} 5 f = \textcolor{w h i t e}{1} 5 f \leftarrow \text{ largest - smallest: next use the 5 & 5}$

$5 f - 5 f = 0 \leftarrow \text{ we have to stop at this point}$

This system is stating that the $G C F = 5 f = 5 \times 8 = 40$
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$\textcolor{b r o w n}{\text{Numeric equivalent}}$

$160 - 120 = 40 \text{ .......} \to \textcolor{w h i t e}{.} 5 f \to \textcolor{w h i t e}{.} 5 \times 8 = 40$
$120 - 40 = 80 \text{ .........} \to 10 f \to 10 \times 8 = 80$
$80 - 40 = 40 \text{ ...........} \to \textcolor{w h i t e}{.} 5 f \to \textcolor{w h i t e}{.} 5 \times 8 = 40$
$40 - 40 = 0$

$G C F = 40$
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$\textcolor{b l u e}{\text{Using prime factor trees}}$

$G C F = 2 \times 2 \times 2 \times 5 = 40$