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 ?
 A: $$45^{n+2}-45^n=45^n(45^2-1)=45^n(45+1)(45-1)\equiv0\pmod{100}$$ 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 $n\ge2$
A: If we take $n=4A+1=10B+5$ where $A,B$ are arbitrary integers
$4A+1=10B+5\iff4(A-1)=10B\iff2(A-1)=5B\implies\dfrac{5B}2=A-1$ which is an integer
As $(2,5)=1,2$ must divide $B\implies B=2C$(say) for some integer $C$
$\implies n=20C+5$
observe that $(10B+5)^2=100(B^2+B)=25\equiv25\pmod{100} \ \ \ \  (1)$
As $((10B+5)^{n+2},100)=25$ for $n\ge0$ let us consider $(20C+5)^n\pmod{\dfrac{100}{25}}$
As $20c+5\equiv1\pmod4\implies(20c+5)^n\equiv1^n\equiv1$
As $a\equiv b\pmod m\implies a\cdot c\equiv b\cdot c\pmod{m\cdot c}$
$\implies(20C+5)^n(20C+5)^2\equiv1\cdot(20C+5)^2\pmod{4(20C+5)^2}$
Using $(1),25|(20C+5)^2\implies100|4(20C+5)^2,$
$(20C+5)^{n+2}\equiv(20C+5)^2\pmod{100}$   for $n\ge0$
But by $(1),(20C+5)^2\equiv25\pmod{100}$
A: HINT:
Suppose $$(100m+45)^n=100k+25,$$ for $n\ge 2$ then $$(100m+45)^{n+1}=(100m+45)(100k+25)=100k'+25.$$
