Huge power

$5603^{1243}\equiv 3^{1243} \pmod {100}\\ 3^{\lambda (100)}=3^{20}\equiv 1\pmod{100}\\ \implies\boxed{1243 \pmod {20}}\\ 1243\equiv 3\pmod {20}\\ \implies 5603^{1243}\equiv 3^{1243}\equiv 3^3\equiv \boxed{27}\pmod{100}$

$\begin{aligned} 5603^{1243} & \equiv (\color{#3D99F6}{5600} + 3)^{1243} \text{ (mod 100)} \quad \quad \small \color{#3D99F6}{\text{Since }5600 \equiv 0 \text{ (mod 100)}} \\ & \equiv \color{#3D99F6}{3}^{1243} \text{ (mod } \color{#3D99F6}{100}) \quad \quad \quad \quad \quad \quad \small \color{#3D99F6}{\text{As 3 and 100 are coprimes, we can use Euler totient theorem.}} \\ & \equiv 3^{1243 \text{ mod } \color{#3D99F6}{\phi(100)}} \text{ (mod 100)} \quad \quad \ \small \color{#3D99F6}{\text{where }\phi (\cdot) \text{ is the totient function.}} \\ & \equiv 3^{1243 \text{ mod } \color{#3D99F6}{40}} \text{ (mod 100)} \\ & \equiv 3^3 \text{ (mod 100)} \\ & \equiv \boxed{27} \text{ (mod 100)} \end{aligned}$

The last 2 digits are 03, $3^{4k}$ has unit digit as 1, thus $3^{1243}$ would have it as 7. Since its followed by a zero, I figured the last 2 digits would just be $03^3=27$ .

P.S- Please let me know if this method is right or wrong.

To simplify, I will first find the value of $x$ in the following:

$\large 5603^x \equiv 1 \pmod {100}$

Using Euler's totient theorem:

$\large \text{Since gcd (100, 5603) = 1, } x = \phi (100) = 100 (1 - \dfrac{1}{2}) (1 - \dfrac{1}{5} ) = 40$

Then using basic modulo arithmetic we can find the answer:

$\large 5603^{40} \equiv 1 \pmod {100} \\ \large \therefore 5603^{1240} \equiv 1 \pmod {100} .................\color{#3D99F6} \text{ Multiply both sides by }31 \\ \large 5603^{1240} \times 5603^3 \equiv 1 \times 3^3 \pmod {100} ................. \color{#3D99F6} \text{ Note that } 5603 \equiv 3 \pmod{100} \\ \large \implies 5603^{123} \equiv 27 \pmod {100} .................\color{#3D99F6} \text{ So 27 is the answer}$