What is the remainder when $25^{889}$ is divided by 99? What is the remainder when $25^{889}$ is divided by 99 ?
$25^3$ divided by $99$ gives $26$ as a remainder.
$25*(25^3)$ divided by $99$ gives (remainder when $25*26$ is divided by $99$) as a remainder.
i.e. $25*(25^3)$ divided by $99$ gives $56$ as a remainder.
$(25^3)*(25^3)$ divided by $99$ gives (remainder when $26*26$ is divided by $99$) as a remainder.
i.e. $(25^3)*(25^3)$ divided by $99$ gives $82$ as a remainder.

$\rm mod\ 9\!:\ 25^3\equiv (-2)^3\equiv -8\equiv 1\:\Rightarrow\:n = 25^{889}\equiv 25^{889\ mod \ 3}\equiv 25\equiv \color{#0A0}{-2}$  
$\rm mod\ 11\!:\ 25^5\equiv 5^{10}\equiv 1\:\Rightarrow\:n = 25^{889}\equiv 25^{889\ mod\ 5}\equiv 25^4 \equiv 3^4\equiv (-2)^2\equiv \color{#C00}4$
$\rm mod\ 9\!:\ \color{#0A0}{{-}2} \equiv n\equiv \color{#C00}4\!+\!11k\equiv 4\!+\!2k\:\Rightarrow\:k\equiv -3\equiv\color{blue}{6}\:\Rightarrow\:n = 4\!+\!11(\color{blue}{6}\!+9\,j) = 70+99\,j$
