Elementary Number Theory – Find the Last Two Digits of 7^81

Question

I came across the following problem and do not know how to tackle it.

Find the last two digits of $ 7^{81} ?$

Can someone point me in the right direction? Thanks in advance for your time.

Answer 1

Last $2$ digits of powers of $7$:

$7^1 = 07$

$7^2 = 49$

$7^3 = 43$

$7^4 = 01$

..................................

$7^5 = 07$

$7^6 = 49$

$7^7 = 43$

$7^8 = 01$

So the last two digits repeat in cycles of $4$.

Hence $7^{81}$ has the last two digits same as that of $7^1$. So answer is : $07$