View source for Euler's Totient Theorem Problem 2 Solution

==Problem==
(BorealBear) Find the last two digits of <math> 3^{3^{3^{3}}} </math>. 

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

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

Link back to [[Euler's Totient Theorem]].