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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.

Bill Dubuque
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Dom
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    Presumably one wants the least positive representative of the inverse, so you should reduce $11^{11}\pmod {26}$. Note that it is somewhat simpler to work $\pmod {13}$ and use the Chinese Remainder Theorem. Smaller moduli tend ot be easier to work with numerically. – lulu Nov 25 '23 at 10:57
  • Also worth noting: with numbers as small as this, just searching for the inverse is very fast and reliable. – lulu Nov 25 '23 at 10:57
  • $26=2\cdot 13$, so you can use this. See also this duplicate. So $19$ is the inverse. – Dietrich Burde Nov 25 '23 at 11:00
  • That’s Euler’s theorem. Not that it matters much. – lhf Nov 25 '23 at 11:42

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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:

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

DonAntonio
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