Step 1

The given linear congruence equation is

\(\displaystyle{25}{x}\equiv{15}{\left({b}\text{mod}{29}\right)}\)

Here, a=25, b=15, m=29,

And,

gcd(a,m)=gcd(25,29)=1

Hence the congruence has 1 incongruent solution which is given by solving the corresponding Diophantine equation

\(\displaystyle{a}{x}+{b}{y}={m}\Rightarrow{25}{x}+{15}{y}={29}\)

Step 2

Let us check the value of x=11.

That is, at x =11,

\(\displaystyle{11}\times{25}+{15}={275}+{15}={290}\)

which is \(\displaystyle{0}{b}\text{mod}{29}\).

Hence x=11 is the solution of given equation.

The given linear congruence equation is

\(\displaystyle{25}{x}\equiv{15}{\left({b}\text{mod}{29}\right)}\)

Here, a=25, b=15, m=29,

And,

gcd(a,m)=gcd(25,29)=1

Hence the congruence has 1 incongruent solution which is given by solving the corresponding Diophantine equation

\(\displaystyle{a}{x}+{b}{y}={m}\Rightarrow{25}{x}+{15}{y}={29}\)

Step 2

Let us check the value of x=11.

That is, at x =11,

\(\displaystyle{11}\times{25}+{15}={275}+{15}={290}\)

which is \(\displaystyle{0}{b}\text{mod}{29}\).

Hence x=11 is the solution of given equation.