Using the Euclidean Algorithm to solve 13 in Z30

30 = 2x13+4

13= 3x4+1

Eliminating multiples of four

1= 13-3x4

= 13-3(30-2x13)

= 7x13-3x30 ----How was this achieved?

Thanks

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- Aug 13th 2010, 10:16 AMmark090480Euclidean Algorithm
Using the Euclidean Algorithm to solve 13 in Z30

30 = 2x13+4

13= 3x4+1

Eliminating multiples of four

1= 13-3x4

= 13-3(30-2x13)

= 7x13-3x30 ----**How was this achieved?**

Thanks - Aug 14th 2010, 03:48 AMDefunkt
$\displaystyle 13-3(30-2 \cdot 13) = 13-(3 \cdot 30 - 6 \cdot 13) = 13 - 3 \cdot 30 + 6 \cdot 13 = (6+1) \cdot 13 - 3 \cdot 30 = 7 \cdot 13 - 3 \cdot 30$

- Aug 14th 2010, 05:17 AMHallsofIvy
- Sep 14th 2010, 06:53 AMmark090480
Could you explain what is going on here? I don't understand why we do it in that way. My maths books just tells me how and not why. Thanks.