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.
 A: Note that the first five powers of 21 end in 21, 41, 61, 81, and 01, respectively. So the pattern repeats every five terms, and the result is therefore 81.
A: 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 $6 \pmod{25} \implies 81 \pmod{100}$.
A: Applying the Lemma below, with $\,u(a) = $ units digit of $a,\,$ and $\,t(a) =$ tens digit,  we obtain
$\ \ u(21^{1234}) = 1,\,\ t(21^{1234}) = u(123\color{#c00}4)\,t(\color{#0a0}21) = \color{#c00}4\cdot\color{#0a0}2 = \color{#f0f}8,\ $ so $\, \ \bbox[6px,border:1px solid red]{\color{#0a0}21^{\large 123\color{#c00}4}\equiv \color{#f0f}81\pmod{\!100}}$
Lemma $\ \ u(a)=1\ \Rightarrow\ u(a^n)=1,\ \ t(a^n) = u(n)\,t(a)\bmod 10\ \ $ [Tens Logarithm Law]
Proof $\,\ a = 1\!+\!10k\,\Rightarrow\, a^n = (1\!+\!10k)^n = 1+10nk + 100(\cdots)\,$ by the Binomial Theorem.
This implies  $\ u(a^n)=1,\ $ and $\,\bmod 10\!:\,\ t(a^n) \equiv  nk\equiv n\,t(a)\equiv u(n)\,t(a)$
Remark $ $ Notice in particular: $\bmod 10\!:\ t(a^n) \equiv n\, t(a),\,$ thus the name "tens logarithm"
See also here.
A: $3^4< 100 < 3^5 = 243 \equiv 43 \pmod{100}$
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 $21^{1234} = 3^{12347}7^{1234} \equiv 69*49 \equiv (70 -1)(50-1) \equiv 3500 -70-50 + 1\equiv 81\pmod{100}$.
