It's My Birth Date

Let $N=12^{8^{1990}}$ . We need to find $N \bmod 100$ . Using Chinese remainder theorem (CRT) and considering 4 and 25, the factors of 100, separately,

Factor 4: $N \equiv 12^{8^{1990}} \equiv 0 \text{ (mod 4)}$ .

Factor 25: Since $\gcd (12, 25) = 1$ , we can apply the Euler's theorem and Euler totient function $\phi (25) = 25 \times \dfrac 45 = 20$ . Then we have:

$\begin{aligned} N & \equiv 12^{8^{1990} \bmod \phi(25)} \text{ (mod 25)} \\ & \equiv 12^{\color{#3D99F6} 8^{1990} \bmod 20} \text{ (mod 25)} & \small \color{#3D99F6} \text{By CRT (see note)} \\ & \equiv 12^{\color{#3D99F6} 4} \text{ (mod 25)} \\ & \equiv (10+2)^4 \text{ (mod 25)} \\ & \equiv 10\times 2^3+2^4 \text{ (mod 25)} \\ & \equiv 86 \equiv 11 \text{ (mod 25)} \end{aligned}$

This implies that $N \equiv 25n + 11$ , where $n$ is an integer. Then $25n+11 \equiv 0 \text{ (mod 4)}$ $\implies n \equiv 1$ and $N \equiv 25+11 \equiv \boxed {36} \text{ (mod 100)}$ .

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Note: Using CRT on $8^{1990} \bmod 20$

$\begin{cases} 8^{1990} \equiv 0 \text{ (mod 4)} \\ 8^{1990} \equiv 8^{1990 \bmod \phi(5)} \equiv 8^{1990 \bmod 4} \equiv 8^2 \equiv 64 \equiv 4 \text{ (mod 5)} \end{cases}$

$\implies 8^{1990} \equiv 5n + 4 \equiv 0 \text{ (mod 4)}$ $\implies n = 0$ and $8^{1990} \equiv 4 \text{ (mod 20)}$