In a previous part of the question, I am asked to find $11^{-1} \mod 40$. I've done that, the answer's $11$. The question continues:

find $x$ where $x^{11} \mod 41 = 10$ showing how you could get the final answer with a calculator that can manage a maximum of $9$ digits.

I'm pretty sure I need Fermat's little theorem, but I'm not sure.

  • $\begingroup$ Combine Mark's answer with Sonnhard's hint. The latter (suitably interpreted) gives you some information on the order of $10$ modulo $41$, in other words, gives you "small" exponents $m$ such that $10^m\equiv1\pmod{41}$. $\endgroup$ – Jyrki Lahtonen Jan 6 '15 at 20:45

Hint: Little Fermat tells us that $x^{40}\equiv 1 \mod 41$ for $x$ not divisible by $41$ and this means that $x^{40r+1}\equiv x \mod 41$. Use what you already know to sort out the exponent correctly. Then manage the arithmetic so your calculator can handle it.


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