Find the last two digits of $47^{89}$ Find the last two digits of the number $47^{89}$
I applied the concept of cyclicity
$47\cdot 47^{88}$
I basically divided the power by 4 and then calculated $7^4=2401$ and multplied it with 47 which gave me the answer $47$ but the actual answer is $67$. How?

$(50-3)^2\equiv10-1\pmod{100}$
$\implies47^{4n+1}=47(47^2)^{2n}\equiv47(10-1)^{2n}$
Now $\displaystyle(10-1)^{2n}=(-1+10)^{2n}\equiv1-\binom{2n}110\pmod{100}\equiv1+80n$
Here $4n+1=89\iff n=?$
Can you take it from here?
