Last 2 digits of $2345^{369}$ http://i.stack.imgur.com/hte3J.jpg
This webpage says last 2 digits of $2345^{369}$ is $75$.
But considering only last 2 digits:
$45^1 = 45$
$45^2 = 25$
$45^3 = 25$
The last 2 digits are always 25. Aren't they ?

$$45^{n+2}-45^n=45^n(45^2-1)=45^n(45+1)(45-1)\equiv0\pmod{100}$$ for $n\ge2$ as $25|45^n$ and $4|(45-1)$
$\implies45^{n+2}\equiv45^n\pmod{100}$
Now $45^2=(50-5)^2=50^2-2\cdot50\cdot5+5^2\equiv5^2\pmod{100}$
and $45^3\equiv45\cdot25\equiv(50-5)(20+5)\equiv50-25\pmod{100}$

Alternatively, $$45^n=(5+40)^n\equiv5^n+\binom n15^{n-1}40\pmod{100}$$
For $n-1\ge1\iff n\ge2,$  $$45^n\equiv5^n\pmod{100}$$
Like the 1st method, we can prove $5^n\equiv5^2\pmod{100}$ for $n\ge2$
