Find the remainder when $787^{777}$ is divided by $100$? 
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.
 A: 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 $01$
A: $787\equiv-13\pmod{100}$
$\implies787^{777}\equiv(-13)^{777}\equiv-13^{777}$
Now $13^2=170-1\implies13^{777}=13(-1+170)^{388}$
and $(-1+170)^{388}\equiv(-1)^{388}+\binom{388}1(-1)^{387}170\pmod{100}$
$\equiv1-388\cdot170$
Again as $388\cdot17\equiv6\pmod{10},388\cdot170\equiv60\pmod{100}$
Hope you can take it from here!
A: We have $\pmod{100}$
$$787^{777}\equiv 87^{777}$$
And we have the following factorisations
$$87=3\times 29$$
$$777=3\times 7\times 37$$
$$87^3=658503\equiv 3\pmod{100}$$
And so
$$787^{777}\equiv 3^{7\times 37}\pmod{100}$$
Now we have $3^7=2187\equiv 87$ and so
$$787^{777}\equiv (3\times 29)^{37}\pmod{100}$$
So we're now left with $3^{37}$ and $29^{37}$ $\pmod{100}$.
A direct computation shows that $3^{15}=14348907\equiv 7\pmod{100}$ and $3^7=2187\equiv 87$ so
$$3^{37}\equiv 7\times 7\times 87=4263\equiv 63\pmod{100}$$
Another direct (and painful) computation shows that $29^7=17249876309\equiv 9$ and $29^{10}=420707233300201\equiv 1$ and so
$$29^{37}\equiv 9\pmod{100}$$
And now putting the two results together
$$787^{777}\equiv 9\times 63=567\equiv 67\pmod{100}$$
