Question

Find the last digit of 3^3^100.

Answer 1

this would be solved through cyclicity u must have learnt about that ?

Answer 2

We'll solve by using Euler's Theorem, a nice extension of Fermat's little theorem. We work mod 10, since that just gives the value of the ones place. For example, 2984 mod 10 = 4

We must first note that 3 and 10 are relatively prime and hence Euler's Theorem applies. Also note that \[\phi(10)=4\]and
\[3^{3^{100}}=3^{300}=(3^{4})^{75}\]

Now apply Euler's Theorem: \[3^{4}\] is congruent to 1, so \[(3^{4})^{75}=1^{75}=1\] mod 10

Hence the last digit is 1.