A number theory problem by abhishek alva

We need to solve for $2^{1999} \pmod{100}$ . Let us consider $2^{1999} \pmod{4}$ and $2^{1999} \pmod{25}$ separately.

$\begin{aligned} 2^{1999} & \equiv 2\cdot 4^{999} \equiv 0 \pmod{4} \end{aligned}$

$\begin{aligned} 2^{1999} & \equiv 2^{\color{#3D99F6}{1999 \mod \phi(25)}} \pmod{25} \quad \quad \small \color{#3D99F6}{\text{Since }\gcd (2,25)=1 \text{, we can use Euler's theorem.}} \\ & \equiv 2^{\color{#3D99F6}{1999 \mod 20}} \pmod{25} \\ & \equiv 2^{19} \pmod{25} \\ & \equiv 1024 \cdot 512 \pmod{25} \\ & \equiv (-1) \cdot (12) \pmod{25} \\ & \equiv 88 \pmod{25} \end{aligned}$

Since $88 \equiv \begin{cases} 0 \pmod{4} \\ 88 \pmod{25} \end{cases} \implies 2^{1999} \equiv \boxed{88} \pmod{100}$

The number given : 2^{1999} This number can also be written as → (2^{10})^{199}^{9}. Point to be noted that 2^{10}^{a} is of the form you find two conditions arises if a is even then then the two units place will be 76 but if its odd then the two units place will be 24. Hence the solution becomes → 24 ×2^{9}=512×24=12288 hence answer is 88.... Q.E.D.