Exponential G.P

Let's call The given number as $\mathfrak{S}$

Method using the Chinese remainder theorem:

Please refer to this wiki if you're unfamiliar with how to calculate a modulo using the CRT.

Break mod 100 into two parts, two numbers coprime to each other.

$1) \mod 4$

$48 \equiv 0 \mod 4, \therefore \mathfrak{S} \equiv 0 \mod 4$

$2) \mod 25$

$\mathfrak{S} \equiv (25+23)^{12^{6^3}} \equiv 23^{12^{6^3}} \mod 25$

Because 23 and 25 are coprime, we can use Euler's Theorem

$23^{20} \equiv 1 \mod 25 ,\because \phi(25) =20$

$\therefore \mathfrak{S} \equiv -2^{24^{12^{6^3} }\mod 20} \mod 25$

Now, $24^{12^{6^3}} \equiv 4^{12^{6^3}} \mod 20$

Observe that: $4^{2n} \equiv 16 \mod 20$

$\because 24^{12^{6^3}} \equiv 0 \mod 2, \therefore 24^{12^{6^3}} \equiv 16 \mod 20$

$\therefore \mathfrak{S} \equiv -2^{16} \equiv 4^8 \equiv 16^4 \equiv 256^2 \equiv 6^2 \equiv 11 \mod 25$

Now, we apply the CRT.

$n_1 = 4, y_1 = 25 ,z_1 =1, a_1 =0$

$n_2 =25, y_2 =4 , z_2 = 19, a_2 = 11 \text{(As calculated above)}$

$\therefore \mathfrak{S} \equiv \Sigma^2_{i=1} a_iy_iz_i = 0+76 \times 11 \equiv 836 \equiv 36 \mod 100$

$\text{Answer} = 36$

Q.E.D.

Let the number be $\large 48^\alpha = 48^{24^\beta} = 48^{24^{12^\gamma}}$ . We need to find $48^\alpha \text{ mod }100$ . Let us consider $n \text{ mod }4$ and $n \text{ mod }25$ separately.

$\begin{aligned} 48^\alpha & \equiv 0 \text{ (mod 4)} \\ 48^\alpha & \equiv 48^{\color{#3D99F6}{24^\beta \text{ mod }\phi (25)}} \text{ (mod 25)} & \small \color{#3D99F6}{\text{As }\gcd (48,25) = 1, \text{ Euler's theorem applies.}} \\ & \equiv 48^{\color{#3D99F6}{24^\beta \text{ mod }20}} \text{ (mod 25)} \end{aligned}$

Now consider $\color{#3D99F6}{24^\beta \text{ mod }20}$ .

$\begin{aligned} 24^\beta & \equiv 0 \text{ (mod 4)} \\ 24^\beta & \equiv 24^{\color{#3D99F6}{12^\gamma \text{ mod }\phi (5)}} \text{ (mod 5)} & \small \color{#3D99F6}{\text{As }\gcd (24,5) = 1, \text{ Euler's theorem applies.}} \\ & \equiv 24^{\color{#3D99F6}{12^\gamma \text{ mod }4}} \text{ (mod 5)} \\ & \equiv 24^{0} \text{ (mod 5)} \\ & \equiv 1 \text{ (mod 5)} \\ \implies 24^\beta & \equiv 96 \text{ (mod 20)} & \small \color{#3D99F6}{\text{Since }4\times 24 = 96 \equiv 0 \text{ (mod 4)} \equiv 1 \text{ (mod 5)}} \\ & \equiv 16 \text{ (mod 20)} \end{aligned}$

$\begin{aligned} \implies 48^\alpha & \equiv 48^{\color{#3D99F6}{24^\beta \text{ mod }20}} \text{ (mod 25)} \\ & \equiv 48^{16} \text{ (mod 25)} \\ & \equiv (-2)^{16} \text{ (mod 25)} \\ & \equiv 2^{16} \text{ (mod 25)} \\ & \equiv 1024 \cdot 64 \text{ (mod 25)} \\ & \equiv -14 \text{ (mod 25)} \\ & \equiv 11 \text{ (mod 25)} \end{aligned}$

$\begin{aligned} \implies 48^\alpha & \equiv 32 \times 48 \text{ (mod 100)} \\ & \equiv 1536 \text{ (mod 100)} & \small \color{#3D99F6}{\text{Since } 1536 \equiv 0 \text{ (mod 4)} \equiv 11 \text{ (mod 25)}}\\ & \equiv \boxed{36} \text{ (mod 100)} \end{aligned}$