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.
 A: First, as $1997\equiv -3\mod 100$, $1997^{1998}\equiv(-3)^{1998}=9^{999}\mod 100$.
Second,   consider the set $\;\{9^n\bmod 100\mid n\in\mathbf N\}$; this is a finite set since $\lvert\mathbf Z/\mathbf 100\mathbf Z\rvert =100$. Hence there exist positive integers $a,k$ such that $9^a=9^{a+k}$.
Furthermore, as $9$ (or $1997$) is coprime to $100$, it is invertible modulo $100$. There results the congruence $9^a\equiv 9^a\cdot9^k$ can be simplified to $9^k=1$. 
The same is true for any number coprime to $100$. The value of $k$ varies with the number, but one proves it is always a divisor of $40$ (the cardinal of the set of numbers less than $100$ that are coprime to $100$).
As a consequence, $\;9^{999}\equiv9^{999\bmod 40}\equiv9^{-1}$.
Now $100=11\cdot9+1$, hence $9^{-1}\equiv -11\equiv \color{red}{89\mod 100}$.
Why $a^{40}\equiv1\mod100$ for all $a$ coprime to $100:$
The congruence classes of these elements are the invertible elements of the  ring $\mathbf Z/100\mathbf Z$. As
$$\mathbf Z/100\mathbf Z\simeq \mathbf Z/4\mathbf Z\times \mathbf Z/25\mathbf Z$$
by the Chinese remainder theorem, it suffices to count the  invertible elements of both factors.
For the first factor, it is readily done: there are two invertible elements, $1$ and $3\bmod 4$. For the second factor, observe the non-invertible elements are the multiples of $5$ modulo $25$, which are $5$, namely $0,5,10,15$ and $20\bmod 25$. Hence $\mathbf Z/25\mathbf Z$ has $25-5$ invertible elements.
Combining both, we obtain $2\times 20=40$ invertible elements modulo $100$.
Now these invertible elements are a (multiplicative) group of order $40$, and by Lagrange's theorem, the order of each element is a divisor of $40$.
A: Using Euler's totient function should save you a considerable amount of time. I assume from your comment that you don't mind if we'll explain a solution involving Euler's totient function, so I'll write one down below. Please comment if something is not clear.
Using properties 2.1.1 and 2.1.2 mentioned in this Wikipedia article, we get that
$\varphi(100)=\varphi(2^2\cdot 5^2)=\varphi(2^2)\cdot\varphi(5^2)=(2^2-2^1)\cdot(5^2-5^1)=40$.
By Euler's theorem (and using the fact that $1997$ is not divisible by $2$ or $5$), we get that $1997^{\varphi(100)}=1997^{40}\equiv1\pmod{100}$.
Notice that $1998=2000-2=50\cdot 40-2$ and $1997\equiv -3 \pmod{100}$.
Overall we get that $$1997^{40}\equiv (1997^{\varphi(100)})^{50}\cdot (-3)^{-2}\equiv 1\cdot 9^{-1}\equiv 89\pmod{100}.$$
In the last step I assumed that you know how to calculate a modular inverse (if not, that's something you should take care of first, unless you intend using a calculator).
A: $$1997\equiv-3\pmod{100}\implies1997^{4n+2}\equiv(-3)^{4n+2}\equiv9^{2n+1}$$
Now $\displaystyle9^{2n+1}=(10-1)^{2n+1}=-(1-10)^{2n+1}\equiv-1+\binom{2n+1}110\pmod{100}\equiv20n+9$
Here $4n+2=1998\iff n=499\equiv4\pmod5\implies20n\equiv20\cdot4\pmod{20\cdot5}$
A: 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 $1997^{1998} \equiv 1997^{60×33+18} \equiv 1997^{18} \equiv 97^{16} × 97^2 \equiv 97^{16} × (-3)^2 \equiv 21 × 9 \equiv 189 \equiv 89$.
