Using Euler's totient theorem to compute $11^{-1}$ mod $26$

Question

I tried to solve this modular inverse equation: $11^{-1}$ mod $26$ with Fermat's little theorem: $a^{\phi (m) - 1} * a \equiv 1$ mod $m$ and reached the following solution:

$11^{\phi (26)-1} * 11 \equiv 1$ mod $26$

$\phi (26) = 12$

$11^{11} * 11 \equiv 1$ mod $26$

Resulting in: $11^{12} \equiv 1$ mod $26$ with the modular inverse of $11$ being $11^{11}$.

Now I'm wondering if my calculations are correct or if there is a different approach and if this generally works this way.

Answer 1

When the numbers are fairly small one can take advantage of a some basic arithmetic.

Since $\;11\cdot7=77=-1\pmod{26}\;$, because $\;26\cdot3=78\;$ , we get:

$$11\cdot(-7)=1\pmod {26}\,,\,\text{and since}\;-7=19\pmod{26}\,,\,\text{we get}\;11^{-1}=19\pmod{26}$$