Find the last two digits in the decimal representation (base 10) of 17^20 My attempt is in this image. But I want the answer to this question by using modulo congruence method since the method used by me involves tedious calculations.

 A: Working modulo $100$,
$17^{20} \equiv 289^{10} \equiv (-11)^{10} \equiv (10+1)^{10} \overset{binom}{\equiv} \binom {10}1 10^1\cdot 1^9 + 1 \equiv 1 \pmod{100}$
So the last two digits are $01$.
A: Euler $\phi(5^2)=5\cdot4=20$ so $\,17^{20}\!\equiv 1\pmod{5^2}.\,$ Also  $\,17^{20}\equiv 1^{20}\equiv 1\pmod 4,\ $ therefore $\,4,25\mid 17^{20}\!-1\,\Rightarrow\, 4\cdot 25\mid 17^{20}-1$
A: Oh, just to be different.
$17^{5} \mod 100 \equiv$
$(20 - 3)^5 \mod 100 \equiv$
$\sum_{k=0}^5 {5 \choose k} 20^k*(-3)^{5-k} \mod 100 \equiv$ [all but the last term is divisible by $20*5 = 100$]
$(-3)^{5} \mod 100$
So 
$17^{20} \mod 100 \equiv$
$(-3)^{20} \mod 100 \equiv$
$9^{10} \mod 100 \equiv$
$(10 - 1)^{10} \mod 100 \equiv$
$\sum_{k=0}^{10}{10 \choose k}10^k(-1)^{10 - k} \mod 100 \equiv$ [all but the last term is divisible by $10*10 = 100$]
$\equiv 1 \mod 100$.
So last two digits are $01$.
....
So... tedious calculations need not be that tedious.
A: The units digit of $17^1$ is equal to $7$.
The units digit of $17^2$ is equal to $9$.
The units digit of $17^3$ is equal to $3$.
The units digit of $17^4$ is equal to $1$.
Calculating,
$\quad 17^4 = (17^2)(17^2) = (289)(289) \equiv (-11)(-11) \equiv 21 \pmod{100}$
Continuing,
$\quad 21^2 \equiv 41 \pmod{100}$
$\quad 21^3 \equiv 61 \pmod{100}$
$\quad 21^4 \equiv 81 \pmod{100}$
$\quad 21^5 \equiv \;\,1 \pmod{100}$
So,
$\quad 17^{20} = (17^4)^5 \equiv 21^5 \equiv 1 \pmod{100}$
