<s>[INST] 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.

 [/INST] 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,
$\quad 17^{20} = (17^4)^5 \equiv 21^5 \equiv 1 \pmod{100}$
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