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$41^5 \equiv 1 \pmod {100}$

$41^{17} = (41^5)^3 \cdot 41^2 \equiv \boxed{81} \pmod {100}$

notice that last digit of 41 is 1. so, any number power 41 will have 1 as last digit.

for second last digit, it is simply the last digit of X(n) = 4*n, where n is the power, since 41 will be multiplied by 1 after each step of multiplication and previous second digit will be added by 4. so for 41^17, X(17) = 68... the second last digit is last digit of 68 which is 8.

this concludes the answer of second last digit is 8 and last digit is 1.

the answer is 81.

$41^{17}( mod 100)≡(40+1)^{17} (mod 100)≡(40^{17}+…+136\times 1600+17\times 40+1) (mod 100)≡281 (mod 100)≡\boxed{81}.$