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?

Your argument would work fine for finding the last two digits of $7^{89}$. But just because $7^4 = 2\,401$ ends in $01$ doesn't mean that $47^4$ will. (In fact, $47^4 = 4\,879\,681$.)
It takes a lot longer for the last digits two of $47^n$ to cycle. You'll at least be able to do calculations with smaller numbers if you think about powers of $47$ separately mod $25$ and mod $4$, then apply the Chinese remainder theorem.
Another approach is to try to compute the last two $47^{89}$ directly by using as few multiplications as possible. For example, if you get the last two digits of $47^{11}$, you can square them three times to get the last two digits of $47^{88}$, and multiply by $47$ again to get to $47^{89}$. (There might be faster ways, too; this is just the first I thought of.)
