Euler's Totient Theorem Problem 2 Solution

Problem

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

Solution

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

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Euler's Totient Theorem Problem 2 Solution

Problem

Solution