201 9 2019 2019^{2019}

What are the last two digits of 201 9 2019 ? 2019^{2019} ?

79 21 61 19

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2 solutions

Chew-Seong Cheong
Dec 22, 2018

201 9 2019 1 9 2019 m o d λ ( 100 ) (mod 100) Since gcd ( 19 , 100 ) = 1 , Euler’s theorem applies 1 9 2019 m o d 20 (mod 100) Carmichael’s lambda function λ ( 100 ) = 20 1 9 19 (mod 100) ( 20 1 ) 19 (mod 100) 380 1 (mod 100) 79 (mod 100) \begin{aligned} 2019^{2019} & \equiv 19^{2019 \bmod \color{#3D99F6} \lambda (100)} \text{ (mod 100)} & \small \color{#3D99F6} \text{Since }\gcd(19,100) = 1 \text{, Euler's theorem applies} \\ & \equiv 19^{2019 \bmod \color{#3D99F6} 20} \text{ (mod 100)} & \small \color{#3D99F6} \text{Carmichael's lambda function } \lambda (100) = 20 \\ & \equiv 19^{19} \text{ (mod 100)} \\ & \equiv (20-1)^{19} \text{ (mod 100)} \\ & \equiv 380 - 1 \text{ (mod 100)} \\ & \equiv \boxed{79} \text{ (mod 100)} \end{aligned}


References:

Michael Leon
Apr 6, 2019

As we know, 2019 is an ood, so we just need to look the ood pattern

we can use the pattern : to found last-2 digits use (mod 100)

201 9 1 19 ( m o d 100 ) 2019^1 \equiv 19 \pmod{100}

201 9 3 59 ( m o d 100 ) 2019^3 \equiv 59 \pmod{100}

201 9 5 99 ( m o d 100 ) 2019^5 \equiv 99 \pmod{100}

201 9 7 39 ( m o d 100 ) 2019^7 \equiv 39 \pmod{100}

201 9 9 79 ( m o d 100 ) 2019^9 \equiv 79 \pmod{100}

201 9 11 19 ( m o d 100 ) 2019^{11} \equiv 19 \pmod{100}

...

So, the pattern repeat every 5 ood number

2019 is 1010th ood, and 1010 is divisible by 5

So, the last 2-digits of 201 9 2019 2019^{2019} same as 201 9 9 = 79 2019^9 = \boxed{79}

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