Step 1

Since gcd(3,11)=1 and again by using Euler's theorem,

\(\displaystyle{3}^{{\phi{\left({11}\right)}}}\equiv{1}{\left(\text{mod}{11}\right)}\)

\(\displaystyle{3}^{{10}}\equiv{1}{\left(\text{mod}{11}\right)}\)

So k can be 10 if no other smaller positive value than 10 satisfies the condition.

Step 2

The possible smaller positive values lesser than 10 satisfy the condition can be the divisors of 10.

Divisors of 10 less than 10 are 1, 2 and 5.

Check the congruences at these values.

\(\displaystyle{3}^{{1}}\equiv{3}{\left(\text{mod}{11}\right)}\)

\(\displaystyle{3}^{{2}}\equiv{9}{\left(\text{mod}{11}\right)}\)

\(\displaystyle{3}^{{4}}\equiv{4}{\left(\text{mod}{11}\right)}\Rightarrow{3}^{{5}}\equiv{1}{\left(\text{mod}{11}\right)}\)

So, 5 satisfies the condition of congruence.

Therefore, k=5.

Since gcd(3,11)=1 and again by using Euler's theorem,

\(\displaystyle{3}^{{\phi{\left({11}\right)}}}\equiv{1}{\left(\text{mod}{11}\right)}\)

\(\displaystyle{3}^{{10}}\equiv{1}{\left(\text{mod}{11}\right)}\)

So k can be 10 if no other smaller positive value than 10 satisfies the condition.

Step 2

The possible smaller positive values lesser than 10 satisfy the condition can be the divisors of 10.

Divisors of 10 less than 10 are 1, 2 and 5.

Check the congruences at these values.

\(\displaystyle{3}^{{1}}\equiv{3}{\left(\text{mod}{11}\right)}\)

\(\displaystyle{3}^{{2}}\equiv{9}{\left(\text{mod}{11}\right)}\)

\(\displaystyle{3}^{{4}}\equiv{4}{\left(\text{mod}{11}\right)}\Rightarrow{3}^{{5}}\equiv{1}{\left(\text{mod}{11}\right)}\)

So, 5 satisfies the condition of congruence.

Therefore, k=5.