Find the remainder when $787^{777}$ is divided by $100$?
MyApproach
$787^{20\times38+17}$=$787^{17}$=I will get the last digit of remainder as 7 but how to calculate tens digit in this question fast using this approach only.
Similarly,Find the remainder when $948^{728}$ is divided by $100$.
On solving I get $948^8$=I will get the last digit of remainder as 7 but how to calculate tens digit in this question fast using this approach only.
Again here how to calculate the other digits fast.
Best Answer
Hint
$787^3 \equiv 03\pmod {100}$ and $3^{20} \equiv 01 \pmod {100}$
Now the problem becomes much simpler. The last two digits of $787^{780}$ are $01$. You can now easily work backwards.
Other Problem
You can tackle it similarly by observing
$948^2 \equiv 04\pmod {100}$ and $4^{16} \equiv 04 \pmod {100}$
(You are not going to be lucky because with an even number, you will never get a $01$)
Edit - Alternately for first problem
You can use Fermat's little theorem, knowing $\phi(100) = 40$.
So any number, relatively prime with $100$, raised to $40$ will give $01$ and hence the last two digits of $787^{780}$ are $01$