Your idea matters-11

$3^{20}\equiv 1(mod 100)$ ,so $3^{10987654320}\equiv 1(mod 100)$ ,since $20|10987654320$

Since $\gcd(3,100)=1$ or 3 and 100 are coprime integers, then Euler's theorem applies. And we note that the Carmichael's lambda function $\lambda(100) = 20$ . Then we have:

$\begin{aligned} 3^{10987654320} & \equiv 3^{10987654320 \bmod \lambda(100)} \pmod {100} \\ & \equiv 3^{10987654320 \bmod 20} \pmod {100} \\ & \equiv 3^0 \equiv 1 \pmod {100} \end{aligned}$

Therefore, $x = \boxed 1$ .

$3^{10987654320}=(10-1)^{5493827160}$ which is congruent to $01$ mod $100$ (using binomial theorem).