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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$\displaystyle 13-3(30-2 \cdot 13) = 13-(3 \cdot 30 - 6 \cdot 13) = 13 - 3 \cdot 30 + 6 \cdot 13 = (6+1) \cdot 13 - 3 \cdot 30 = 7 \cdot 13 - 3 \cdot 30$
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) = 7x13-3x30 ---- How was this achieved? Thanks
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.
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