+65 I believe that your answer would be 25 and a remainder of 8
sincerely ▄︻デ✞☬🇮🇫1️⃣🇳1️⃣🇹¥☬✞══━一
+9519 Note that \(\varphi(13) = 13 - 1 = 12\), where \(\varphi\) denotes the Euler totient function.
Now, for these type of problems, you can take mod 13 of the base and mod \(\varphi(13)\) of the power.
\(\begin{array}{rcl} 333^{333} &\equiv& 8^9 \pmod{13}\\ &=& 2^{27} \pmod{13}\\ &=& 2^3 \pmod{13} \end{array}\)
Note that I used \(8^9 = (2^3)^9 = 2^{27}\) and then took mod \(\varphi(13)\) of the power again.
Now the size of the problem becomes manageable, because you can calculate 2^3, and that's the answer.
