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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.

enter image description here

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    $01$ is the right answer – 2017-01-08
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    I know that. I told I want the solution using modulo operator. – 2017-01-08
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    You can search the questions on MSE for a dozen of similar questions. – 2017-01-08
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    Is MSE maths Stack exchange? – 2017-01-08
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    We expect you to search the site before posting. See the list of "Related" questions in the right margin. There are several very good matches in there such as [this](http://math.stackexchange.com/q/290394/11619). We have also collected many relevant techniques to [this thread](http://math.stackexchange.com/q/81228/11619) – 2017-01-08

3 Answers 3

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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 $01$.

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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 $\,4,25\mid 17^{20}\!-1\,\Rightarrow\, 4\cdot 25\mid 17^{20}-1$

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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.