Euclidean algorithm to find integers $s$ and $t$ such that $sa+tb=1$

Question

$a=19845$, $b=218$

I got this far but am now stuck and don't know what to do.

19845 % 218 = 7
218 % 7 = 1
1 = 218 - (7*31)
7 = 19845 - (218*91)
1 = 218 - ((19845-218*91)*31)

Then from there I do not know how to simplify it to get the values for $s$ and $t$.

score 3 · Accepted Answer · answered Nov 05 '13 at 12:51

Use the distributive law: \begin{align} 1 &= 218 - (19845-218\cdot 91)\cdot 31 \\ &= 218 - 19845\cdot 31 + 218\cdot 91\cdot 31 \\ &= 218 \cdot (1 + 91\cdot 31) + 19845\cdot (-31) \\ &= 218\cdot 2822 + 19845\cdot (-31) \end{align} Thus, $s=-31$ and $t=2822$.

score 2 · Answer 2 · edited Apr 13 '17 at 12:21

2

Alternatively, you can use the Euclid-Wallis Algorithm: $$ \begin{array}{r} &&91&31&7\\\hline 1&0&1&-31&218\\ 0&1&-91&2822&-19845\\ 19845&218&7&1&0\\ \end{array} $$ This says that $$ 2822\cdot218-31\cdot19845=1 $$ The bookkeeping incorporated in that algorithm takes care of the difficulty that often accompanies the Euclidean Algorithm.

edited Apr 13 '17 at 12:21

Community

1

answered Nov 05 '13 at 12:54

robjohn

345,667

Is this just the Extended Euclidean Algorithm written sideways? – Christoph Nov 05 '13 at 13:00
That article seems to have gone under many changes recently, but the algorithm seems to be the same. It is a common algorithm. – robjohn Nov 05 '13 at 13:24

score 0 · Answer 3 · answered Dec 18 '19 at 02:37

0

$ 218t+19845s = 1\iff \bmod 218\!:\ s\equiv\dfrac{1}{19845}\equiv \dfrac{1}7\equiv \dfrac{-217}7\equiv -31$

answered Dec 18 '19 at 02:37

Bill Dubuque

272,048

Euclidean algorithm to find integers $s$ and $t$ such that $sa+tb=1$

3 Answers3