Simple 2018 problem.

What are the last two digits of 201 8 2016 2018^{2016} ?

Hint: Pattern


The answer is 76.

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1 solution

Chew-Seong Cheong
Sep 10, 2018

Relevant wiki: Chinese Remainder Theorem

We need to find 201 8 2016 m o d 100 2018^{2016} \bmod 100 . Since gcd ( 2018 , 100 ) 1 \gcd(2018, 100) \ne 1 , we need to consider the factors 4 and 25 of 100 separately using Chinese remainder theorem.

Factor 4: 201 8 2016 0 (mod 4) 2018^{2016} \equiv 0 \text{ (mod 4)} .

Factor 25:

201 8 2016 ( 2000 + 18 ) 2016 (mod 25) 1 8 2016 (mod 25) ( 25 7 ) 2016 (mod 25) 7 2016 (mod 25) 4 9 1008 (mod 25) ( 50 1 ) 1008 (mod 25) 1 1008 1 (mod 25) \begin{aligned} 2018^{2016} & \equiv (2000+18)^{2016} \text{ (mod 25)} \\ & \equiv 18^{2016} \text{ (mod 25)} \\ & \equiv (25-7)^{2016} \text{ (mod 25)} \\ & \equiv 7^{2016} \text{ (mod 25)} \\ & \equiv 49^{1008} \text{ (mod 25)} \\ & \equiv (50-1)^{1008} \text{ (mod 25)} \\ & \equiv 1^{1008} \equiv 1 \text{ (mod 25)} \end{aligned}

This implies that 201 8 2016 25 n + 1 2018^{2016} \equiv 25n + 1 , where n n is an integer. Then we have:

25 n + 1 0 (mod 4) n + 1 0 (mod 4) n = 3 201 8 2016 25 ( 3 ) + 1 (mod 100) 76 (mod 100) \begin{aligned} 25n + 1 & \equiv 0 \text{ (mod 4)} \\ n + 1 & \equiv 0 \text{ (mod 4)} & \small \color{#3D99F6} \implies n = 3 \\ \implies 2018^{2016} & \equiv 25(3) + 1 \text{ (mod 100)} \\ & \equiv \boxed{76} \text{ (mod 100)} \end{aligned}

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