Find last two digit I have the following task: $1997^{1998} \pmod {100} = ?$ 
How to find it? Could you please, explain to me step by step with?
Can you suggest any solution, without using Euler function? But rather, with binomial theorem.

First of all, only the last two digits of 1997 will have an impact on your answer. This is because $100^n \equiv 0$ and $(100 + m)^n \equiv m^n (mod 100)$.
So we need to find the last two digits of $97^{1998}$. We can proceed by calculating various powers of 97 (mod 100).
$97 \equiv -3$
$97^2 \equiv 9$
$97^4 \equiv 81 \equiv -19$
$97^8 \equiv (-19)^2 = 361 \equiv 61$
$97^{16} \equiv 61^2 = 3600+120+1 \equiv 21$
$97^{32} \equiv 21^2 = 441 \equiv 41$
$97^{64} \equiv 41^2 = 1600+80+1 \equiv 81$.
So $97^{64} \equiv 97^4$ so $97^{60} \equiv 1$, so $1997^{60} \equiv 1$.
So $1997^{1998} \equiv 1997^{60×33+18} \equiv 1997^{18} \equiv 97^{16} × 97^2 \equiv 97^{16} × (-3)^2 \equiv 21 × 9 \equiv 189 \equiv 89$.
