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.
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.