1. Extended Euclidean Algorithm for `11` and `12`

Solution:
`y-=z^(-1)\ ("mod "n)-=11^(-1)\ ("mod "12)-=11\ ("mod "12)`

Extended Euclidean Algorithm

i	Quotient `q=r_1-:r_2`	Remainder `r=r_1-q*r_2`	`s=s_1-q*s_2`	`t=t_1-q*t_2`
1		`r_1=12`	`s_1=1`	`t_1=0`
2		`r_2=11`	`s_2=0`	`t_2=1`
3	`12-:11=1`	`12-1xx11=1`	`1-1xx0=1`	`0-1xx1=-1`
4	`11-:1=11`	`11-11xx1=0`	`0-11xx1=-11`	`1-11xx-1=12`

we get answer by taking the last non-zero row for Remainder `r=1` (gcd), `s=1,t=-1`

Showing the identity, we have `12s+11t = gcd(12,11)=1`

`12(1)+11(-1)=1`

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2. Extended Euclidean Algorithm for `7` and `11`

Solution:
`y-=z^(-1)\ ("mod "n)-=7^(-1)\ ("mod "11)-=8\ ("mod "11)`

Extended Euclidean Algorithm

i	Quotient `q=r_1-:r_2`	Remainder `r=r_1-q*r_2`	`s=s_1-q*s_2`	`t=t_1-q*t_2`
1		`r_1=11`	`s_1=1`	`t_1=0`
2		`r_2=7`	`s_2=0`	`t_2=1`
3	`11-:7=1`	`11-1xx7=4`	`1-1xx0=1`	`0-1xx1=-1`
4	`7-:4=1`	`7-1xx4=3`	`0-1xx1=-1`	`1-1xx-1=2`
5	`4-:3=1`	`4-1xx3=1`	`1-1xx-1=2`	`-1-1xx2=-3`
6	`3-:1=3`	`3-3xx1=0`	`-1-3xx2=-7`	`2-3xx-3=11`

we get answer by taking the last non-zero row for Remainder `r=1` (gcd), `s=2,t=-3`

Showing the identity, we have `11s+7t = gcd(11,7)=1`

`11(2)+7(-3)=1`

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3. Extended Euclidean Algorithm for `3` and `7`

Solution:
`y-=z^(-1)\ ("mod "n)-=3^(-1)\ ("mod "7)-=5\ ("mod "7)`

Extended Euclidean Algorithm

i	Quotient `q=r_1-:r_2`	Remainder `r=r_1-q*r_2`	`s=s_1-q*s_2`	`t=t_1-q*t_2`
1		`r_1=7`	`s_1=1`	`t_1=0`
2		`r_2=3`	`s_2=0`	`t_2=1`
3	`7-:3=2`	`7-2xx3=1`	`1-2xx0=1`	`0-2xx1=-2`
4	`3-:1=3`	`3-3xx1=0`	`0-3xx1=-3`	`1-3xx-2=7`

we get answer by taking the last non-zero row for Remainder `r=1` (gcd), `s=1,t=-2`

Showing the identity, we have `7s+3t = gcd(7,3)=1`

`7(1)+3(-2)=1`

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