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.

Euler’s Number of $99$
= $99.\frac{2}{3}.\frac{10}{11}$
= $60$
From Fermat’s Theorem we know
$25^{60 × K} \mod {99} = 1$ (where K is any natural number)
Note that $25$ and $99$ are co-primes (∵ they don’t have any common factors other than $1$)
Putting $K =15$, we have, $25^{900} \mod {99} = 1$
Let’s assume $25^{899} \mod {99} = R$
∴ $25^{899} = 99N + R$ (where $N$ is a natural number)
$25^{900} \mod {99} = (25 × 25^{899}) \mod {99} = (25 × R) \mod {99} = 1$
we can conclude that $R=4$
