Problems need HardWork 17

Using Euler's theorem,

$\displaystyle a^{ \phi(p) } \equiv 1$ $mod$ $(p)$ when $gcd(a,p) = 1$

$$ $100 = 2^2 \times 5^2$

$$ $\displaystyle \phi(100) = 2^2 \times 5^2 \times (1 - \dfrac{1}{2} ) \times (1 - \dfrac{1}{5}) = 40$

$$ Hence,

$$ $17^{40} \equiv 1 \pmod{100}$

$$ $17^{82} = 17^{80} \times 17^{2} \equiv 17^2 \equiv 289 \equiv 89 \pmod{100}$

$$

$\therefore 17^{82} = 100n + 89$

$$ $\displaystyle 100n = 17^{82} - 89$

$$ Consider the modular relation,

$$ $\displaystyle 100n \equiv 17^{82} - 89 \equiv 17 \times 17^{81} - 89 \equiv 17 \times 4913^{27} - 89 \pmod{1000}$

$$ $\displaystyle \therefore 100n \equiv 17 \times 913^{27} - 89 \equiv 17 \times 497^{9} - 89 \equiv 17 \times 473^{3} - 89 \equiv 17 \times 817 - 89\pmod{1000}$

$\displaystyle \therefore 100n \equiv 17 \times 817 - 89 \equiv 13889 -89 \equiv 13800 \equiv 800 \pmod{1000}$

$$ $\displaystyle \therefore 100n = 1000k + 800$ , where k is an integer. $\therefore n = 10k + 8$

$$ Hence the last digit of n, that is, the quotient is 8.

If you've noticed, starting from $7^5$ , there is a pattern that goes as 8 6 5 for it's third digit. The last digit has a pattern that goes as 7 9 3 1. Their LCM is 12, so every 12 exponents the third digit will be 5. Divide 82 by 12, which results 72 with 10 as a remainder. Divide the remainder by three which results 9 with 1 as a remainder. That would make the answer the next pattern after 5, which is $\boxed{8}$

First I did $17^{82} mod 1000$ to find the last 3 digits ( which turned out to be $17^{82}=......889$

Since dividing by 100 is basically moving the decimal place to the left, we get the quotient as $......8.89$ or simply as $......8+\frac{89}{100}$ with the fractional part as the remainder.

Thus the answer is $8$