Euler's Totient Theorem Problem 2 Solution

Problem

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

Solution

This problem is just asking for $3^{3^{3^{3}}}\pmod{100}$ . We can start by expanding the uppermost exponent, which gives us $3^{3^{27}}$ . Then, since $\phi(100)=40$ , the exponent will be equal to $3^{27}\pmod{40}$ . We can see that $3^4\equiv 81\equiv 1\pmod{40}$ , so the expression simplifies to $3^3\equiv 27\pmod{40}$ .

We're now left with finding the last two digits of $3^{27}$ . To do this, we use Chinese Remainder Theorem. We find that it is $3$ mod $4$ and $12$ mod $25.$ From here, we use guess+check to get $\boxed{87}$ . -BorealBear

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Euler's Totient Theorem Problem 2 Solution

Problem

Solution