# Euclidean Algorithm

• Aug 13th 2010, 11:16 AM
mark090480
Euclidean 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)

Thanks
• Aug 14th 2010, 04:48 AM
Defunkt
$13-3(30-2 \cdot 13) = 13-(3 \cdot 30 - 6 \cdot 13) = 13 - 3 \cdot 30 + 6 \cdot 13 = (6+1) \cdot 13 - 3 \cdot 30 = 7 \cdot 13 - 3 \cdot 30$
• Aug 14th 2010, 06:17 AM
HallsofIvy
Quote:

Originally Posted by mark090480
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)

= 13- 3(30)- 3(-2(13))= 13- 3(30)+ 6(13)= 7(13)- 3(30)

Quote:

Thanks
• Sep 14th 2010, 07:53 AM
mark090480
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.