Last two digits.

Algebra Level 3

What is the last two digits of 201 7 2017 = ? \large 2017^{2017}=?


The answer is 77.

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1 solution

Chew-Seong Cheong
Oct 14, 2017

Relevant wiki: Euler's Theorem

201 7 2017 1 7 2017 (mod 100) 1 7 2017 mod ϕ ( 100 ) (mod 100) Since gcd ( 17 , 100 ) = 1 , Euler’s theorem applies. 1 7 2017 mod 40 (mod 100) Euler’s totient function ϕ ( 100 ) = 40 1 7 17 (mod 100) 17 ( 1 7 2 ) 8 (mod 100) 17 ( 289 ) 8 (mod 100) 17 ( 300 11 ) 8 (mod 100) 17 × 1 1 8 (mod 100) 17 ( 10 + 1 ) 8 (mod 100) 17 ( 8 × 10 + 1 ) (mod 100) ( 10 + 7 ) ( 80 + 1 ) (mod 100) ( 560 + 10 + 7 ) (mod 100) 77 (mod 100) \begin{aligned} 2017^{2017} & \equiv 17^{2017} \text{ (mod 100)} \\ & \equiv 17^{\color{#3D99F6} 2017 \text{ mod }\phi (100)} \text{ (mod 100)} & \small \color{#3D99F6} \text{Since }\gcd(17, 100) = 1 \text{, Euler's theorem applies.} \\ & \equiv 17^{\color{#3D99F6} 2017 \text{ mod }40} \text{ (mod 100)} & \small \color{#3D99F6} \text{Euler's totient function }\phi (100) = 40 \\ & \equiv 17^{\color{#3D99F6}17} \text{ (mod 100)} \\ & \equiv 17\left(17^2\right)^8 \text{ (mod 100)} \\ & \equiv 17\left(289\right)^8 \text{ (mod 100)} \\ & \equiv 17\left(300-11\right)^8 \text{ (mod 100)} \\ & \equiv 17\times 11^8 \text{ (mod 100)} \\ & \equiv 17(10+1)^8 \text{ (mod 100)} \\ & \equiv 17(8\times 10 +1) \text{ (mod 100)} \\ & \equiv (10+7)(80 +1) \text{ (mod 100)} \\ & \equiv (560+10+7) \text{ (mod 100)} \\ & \equiv \boxed{77} \text{ (mod 100)} \end{aligned}

Thank you for sharing your solution.

Hana Wehbi - 3 years, 7 months ago

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