Arghh......REmainders

Well I will go for an explicit solution...

We know that according to fermat's little remainder theorem:

$a^{p-1}$ $\equiv1$ (mod p)

$4566^{22}$ $\equiv1$ (mod 23)

$4566^{22(46)}$ $\equiv1$ (mod 23)

$4566^{1012}$ $\equiv1$ (mod 23)

Now we have $4566^{9}$ remaining with us...

4566= $2 \times3 \times761$

761 $\equiv2$ (mod 23)

$4566^{9}$ \equiv\(2^{9} . $3^{9}$ . $761^{9}$ ) (mod 23)

$4566^{9}$ \equiv\(2^{9} . $3^{9}$ . $2^{9}$ ) (mod 23)

$4566^{9}$ \equiv\(12^{9} ) (mod 23)

thus we need to find y where

$12^{9}$ $\equiv y$ (mod23)

12 $\equiv 12$ (mod23)

$12^{2}$ $\equiv 6$ (mod23)

$12^{3}$ $\equiv 3$ (mod23)

$12^{3}$ X $12^{3}$ X $12^{3}$ $\equiv 3 X 3 X 3$ (mod23)

= $12^{9}$ $\equiv 27$ (mod23)

= $12^{9}$ $\equiv 4$ (mod23)

Therefore,

$4566^{1021}$ $\equiv 4$ (mod23)

4 = $2^{2}$

Implies that a=2

therefore, 100a= $\huge{200}$

I just guess, a^a<23 so a is 1 or 2

a^a cannot be bigger than 23.

$a^{a}$ < 23. So either 1 or 2.