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 too 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 to use it, only elementary methods.

$3^4< 100 < 3^5 = 243 \equiv 43 \pmod{100}$
So $3^{5k}\equiv 43^{k}\pmod {100}$.
$43^2 = (40 + 3)^2 = 1600 + 2*3*40 + 9 \equiv 49 \pmod {100}$
So $3^{10k} \equiv 49^k \pmod {100}$.
$49^2 = (50-1)^2 = 2500 - 100 + 1 \equiv 1 \pmod {100}$.
So $3^{20k}\equiv 1\pmod{100}$ so
$3^{1234} \equiv 3^{14} \equiv 49*3^4 = (50-1)(80+1) \equiv 4000-80 + 50 -1 \equiv -31\equiv 69\pmod{100}$.
Do similar crap for $7$.
$7^2 = 49$ and so $7^4 = (50-1)^2 = 2500- 100 + 1\equiv 1 \pmod {100}$
So $7^{4k}\equiv 1 \pmod {100}$ and $7^{1234} \equiv 7^2 \equiv 49\pmod {100}$.
So $21^{1234} = 3^{12347}7^{1234} \equiv 69*49 \equiv (70 -1)(50-1) \equiv 3500 -70-50 + 1\equiv 81\pmod{100}$.
