elementary number theory - Find $21^{1234}pmod{100}equiv ?$

The I'm having trouble to do this only by hand (no software or calculator). I tried the following:

\begin{align}21^{1234}(\text{mod} \ 100) &= 21^{1000}21^{200}21^{20}21^4(\text{mod} \ 100)\equiv41^{500}41^{100}41^{15}41^2(\text{mod} \ 100)\\

\end{align}

It's not reasonable to continue taking powers of 21, takes to long with pen and paper. Is there a more efficient way?

Yes I know about the Euler theorem and his totient function but please I don't want t use it, only elementary methods.

Answer

Does the binomial theorem count as an elementary method? If so, we can just do $21^{1234} \equiv (20+1)^{1234} \equiv 20^{1234} + \binom{1234}{1} \cdot 20^{1233} + ... + 20\cdot \binom{1234}{1} + 1 \pmod{100}$. Then, if the exponent of $20$ is greater than or equal to $2$, it is divisible by $100$, so we simply get $20 \cdot 1234 + 1 \equiv 81 \pmod{100}$.

Otherwise, just break it down $\pmod{4}$ and $\pmod{25}$, and evaluate the first few terms, which should give you $1 \pmod{4}$ and $6 \pmod{25} \implies 81 \pmod{100}$.