$2019^{2019}$

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

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References:

• Euler's theorem
• Carmichael's lambda function

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)

$2019^1 \equiv 19 \pmod{100}$

$2019^3 \equiv 59 \pmod{100}$

$2019^5 \equiv 99 \pmod{100}$

$2019^7 \equiv 39 \pmod{100}$

$2019^9 \equiv 79 \pmod{100}$

$2019^{11} \equiv 19 \pmod{100}$

...

So, the pattern repeat every 5 ood number

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

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