Euler's Totient Theorem Problem 2 Solution

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(BorealBear) Find the last two digits of $3^{3^{3^{3}}}$.


Finding the last two digits is equivalent to finding $3^{3^{3^{3}}}\pmod{100}$. We can start by expanding the uppermost exponent: $3^{3^{27}}$. Then, since $\phi(100)=40$, the exponent is equal to $3^{27}\pmod{40}$. We see that $3^4=81\equiv1\pmod{40}$, so it simplifies to $3^3={27}\pmod{40}$.

We now just need to find the last two digits of $3^{27}$. Using the Chinese Remainder Theorem, we find that the last two digits are $3\pmod{4}$ and $12\pmod{25}$. We guess and check to get $\boxed{87}$. ~BorealBear

Link back to Euler's Totient Theorem.