Solution

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Find a number $a$ where $0 \le a \lt 73$ and $a \equiv 9^{794}\pmod{73}$


Note that $73$ is prime and $9 \not\equiv 0\pmod{73}$. Thus, Fermat's Little Theorem applies and guarantees that $9^{72} \equiv 1\pmod{73}$.

Thus $\pmod{73}$ we have, $$\begin{array}{rcl} 9^{794} &=& (9^{72})^{11} \cdot 9^2 \\ &\equiv& 1^{11} \cdot 81 \\ &\equiv& 81 \\ &\equiv& 8 \end{array}$$