Question

Find the inverse of:.$\mathbf{20}\mathbf{}\mathbf{mod}\mathbf{}\mathbf{79}\mathbf{,}\mathbf{}\mathbf{3}\mathbf{}\mathbf{mod}\mathbf{}\mathbf{62}\mathbf{,}\mathbf{}\mathbf{21}\mathbf{}\mathbf{mod}\mathbf{}\mathbf{91}\mathbf{,}\mathbf{5}\mathbf{}\mathbf{mod}\mathbf{}\mathbf{23}$

Short answer

Inverse of the given numbers is obtained.

Step-by-step solution

Inverse of 20 mod 79

$\begin{array}{l}\mathrm{GCD}(20,79)\\ 79=(20\times 3,19)\\ 20=(19\times 1,1)\\ \mathrm{GCD}(20,79)=1\mathrm{so}\mathrm{inverse}\mathrm{of}20\mathrm{mod}79\\ \mathrm{So},\mathrm{By}\mathrm{backward}\mathrm{substitution},\\ 1=20-1\times 19\\ =20-1\times (79-3\times 20)\\ =20-79+3\times 20\\ =4\times 20-1\times 79.\\ \mathrm{So},{20}^{-1}=4\mathrm{mod}79.\end{array}$

Inverse of 3 mod 62

$\begin{array}{l}\mathrm{GCD}(3,62)\\ 62=(3\times 30,2)\\ 3=(2\times 1,1)\\ \mathrm{GCD}(3,62)\mathrm{is}1\mathrm{so},\mathrm{inverse}\mathrm{of}3\mathrm{mod}62\mathrm{is}\\ 1=3-2\\ =3-(62-20\times 3)\\ =3-62+20\times 3\\ =21\times 3-62\times 1\\ \mathrm{So},3^{-1}=21\mathrm{mod}62.\end{array}$

Inverse of 21 mod 91

$\begin{array}{l}\mathrm{GCD}(21,91)\\ 91=21\times 4,7\\ 21=7\times 3,0\\ \mathrm{So},\mathrm{Here}\mathrm{GCD}(21,91)\mathrm{is}0\mathrm{so}\mathrm{inverse}\mathrm{is}\mathrm{not}\mathrm{possible}\mathrm{as}\mathrm{per}\mathrm{EuclidTheorem}\end{array}$

Inverse of 5 mod 23

$\begin{array}{l}\begin{array}{l}\mathrm{GCD}(5,23)\\ 23=(5\times 4,3)\\ 5=(3\times 1,2)\\ 3=(2\times 1,1)\\ \mathrm{GCD}(5,23)\mathrm{is}\mathrm{equal}\mathrm{to}1\mathrm{so}\mathrm{inverse}\mathrm{of}5\mathrm{mod}23\\ 1=3-2\\ =3-(5-3)\\ =3\times 2-5\times 1\\ =(23-4\times 5)\times 2-5\times 1\\ =23\times 2-9\times 5\end{array}\\ \mathrm{So},5-1=-9=14\mathrm{mod}23.\end{array}$

Hence, inverse of given numbers is obtained.