A number theory problem by Kaleem Kħặŋ

Since 6 and 100 are not coprime integers. Let us consider the modulos of $6^{100}$ of 4 and 25 $(100=4\times 25)$ separately.

$\begin{aligned} 6^{100} & \equiv (36)^{50} \equiv 0 \text{ (mod 4)} \end{aligned}$

$\begin{aligned} 6^{100} & \equiv 6^{\color{#3D99F6}100 \text{ mod }\phi (25)} \text{ (mod 25)} & \small \color{#3D99F6} \text{Since }\gcd(6,25)=1 \text{, Euler's theorem applies.} \\ & \equiv 6^{\color{#3D99F6}100 \text{ mod }20} \text{ (mod 25)} & \small \color{#3D99F6} \text{Euler's totient function }\phi(25)=20 \\ & \equiv 6^{\color{#3D99F6}0} \text{ (mod 25)} \\ & \equiv 1 \text{ (mod 25)} \end{aligned}$

Let $6^{100} \equiv 25n +1$

$\begin{aligned} \implies 25n+1 & \equiv 0 \text{ (mod 4)} \\ 25(3)+1 & \equiv 76 \equiv 0 \text{ (mod 4)} \\ \implies 6^{100} & \equiv \boxed{76} \text{ (mod 100)} \end{aligned}$

Powers of $6$ , starting with $6^2=36$ , have the two ending digits equal to $36, 16, 96, 76, 56$ and this sequence of five possible endings repeats.

$6^5$ , and all powers of $6$ divisible by $5$ , which includes $6^{100}$ , end in $\boxed{76}$ .