<s>[INST] 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.
 [/INST] Using Carmichael Function,
 $\lambda(99)=$lcm $(\lambda(9),\lambda(11))=$lcm$(3(3-1),10)=30$
So, $5^{30}\equiv1\pmod {99}$
Now, $25^{889}=(5^2)^{889}=5^{1788}$
Also, $1778\equiv 8\pmod {30}\implies 25^{889}=5^{1780}\equiv5^8\pmod {99}$
$5^2=25,5^3=125\equiv26\pmod{99},5^4\equiv26\cdot5\equiv31\pmod{99},$
$5^8=(5^4)^2\equiv(31)^2\pmod{99}\equiv961\equiv-29\pmod{99}$ as $990=99\cdot10$
So, $5^8\equiv-29\pmod{99}\equiv70$
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