A number theory problem by jitendriya praharaj

We need to find $2^{2014} \text{ mod }1000$ . Let us consider $2^{2014} \text{ mod }8$ and $2^{2014} \text{ mod }125$ , since $1000 = 2^3 \cdot 5^3 = 8 \cdot 125$ , separately.

$\begin{aligned} 2^{2014} & \equiv 0 \text{ (mod 8)} \\ \implies 2^{2014} & \equiv 8n \text{ (mod 1000)} \quad ...(1) & \small \color{#3D99F6}{\text{where }n \text{ is a positive integer.}} \end{aligned}$

$\begin{aligned} 2^{2014} & \equiv 2^{2014 \text{ mod }\color{#3D99F6}{\phi(125)}} \text{ (mod 125)} & \small \color{#3D99F6}{\text{As }\gcd (2, 125) = 1 \text{, Euler's theorem applies.}} \\ & \equiv 2^{2014 \text{ mod }\color{#3D99F6}{100}} \text{ (mod 125)} & \small \color{#3D99F6}{\text{Euler totient function }\phi(125) = 100} \\ & \equiv 2^{\color{#3D99F6}{14}} \equiv (2^7)^2 \equiv 128^2 \equiv 3^2 \\ & \equiv 9 \text{ (mod 125)} \end{aligned}$

From $(1)$ , we have

$\begin{aligned} 8n & \equiv 9 \text{ (mod 125)} \\ \implies 8n & = 125m + 9 & \small \color{#3D99F6}{\text{where }m \text{ is a positive integer.}} \\ n & = \frac {125m + 9}8 & \small \color{#3D99F6}{\text{For integer }n, m \text{ must be odd.}} \end{aligned}$

When $m = 3$ , $n = \dfrac {125(3) + 9}8 = \dfrac {375 + 9}8 = \dfrac {384}8 = 48$ .

From $(1): \quad 2^{2014} \equiv 8n \equiv \boxed{384} \text{ (mod 1000)}$ .