To find the hundredth digit of a number, it is sufficient to compute this number modulo 1000. Because 11 and 1000 are coprime, Euler's totient theorem may be applied, so

`11^( phi ( 1000 ) ) = 1 (mod 1000) .`

It is not hard to compute `phi ( 1000 ) , ` it is equal to `1000 ( 1 - 1 / 2 ) ( 1 - 1 / 5 ) = 400 .`

Because of this, `11^400 = 1 (mod 1000) ` and `11^2000 = 1 ( mod 1000 ) , ` and the initial question is simplified to what is `11^22 ( mod 1000 ) .`

Now express `11^22 ` as `( 11^5 )^4 * 121 , ` and because `12^5 = 161051 , ` we know that

`11^22 (mod 1000) = 51^4 * 121 (mod 1000) = ( 51^2 )^2 * 121 (mod 1000) =`

`2601^2 * 121 mod 1000 = 601^2 * 121 mod 1000 = 201 *121 mod 1000 = 321 .`

This way, the answer is 3.