Computing last two digits of $27^{2018}$ For abstract algebra I have to find the last two digits of $27^{2018}$, without the use of a calculator, and as a hint it says you should work in $\mathbb{Z}/100\mathbb{Z}$. 
I thought breaking up the problem into $\mod(100)$ arguments. Thus:
$27^{2}=729\equiv 29 \mod (100)$, and
$27^{4}=(27^{2})^{2} \equiv 29^{2}=861\equiv 61 \mod (100)$ and
$27^{8}=(27^{4})^{2} \equiv 61^{2}=3421 \equiv 21 \mod(100)$ and so on until I would find something that repeated itself. But I've done quite some terms no and I've not seen any iteration yet. So I'm thinking this is the wrong way.
Any suggestions?
 A: $\phi(100)=40$, so we can reduce the exponent mod $40$:
$$
27^{2018}\equiv27^{18}\pmod{100}
$$
Then we can square and multiply:
$$
\begin{align}
27^2&\equiv29\\
27^4&\equiv41\\
27^8&\equiv81\\
27^9&\equiv87\\
27^{18}&\equiv69\pmod{100}
\end{align}
$$
A: $27^{5} = 7 \quad (\mathrm{mod} \text{ } 100)$. 
So $27^{2018} = 27^{403 \times 5 + 3} = 7^{403} \times 27^3 \quad(\mathrm{mod} \text{ } 100)$.  
Now $7^4 = 1 \quad (\mathrm{mod} \text{ } 100)$.  So $27^{2018} = 7^{4 \times 100 + 3} \times 27^3 = 7^3 \times 27^3$.
And $7 \times 27 = 189 = -11 \quad (\mathrm{mod} \text{ } 100)$. So $(7 \times 27)^3 = - 11^3 = 69 \quad (\mathrm{mod} \text{ } 100)$. You get finally $$27^{2018} = 69 \quad (\mathrm{mod} \text{ } 100)$$
A: $\bmod 100\!:\,\ 3^{\large 3\cdot 2018}\!\equiv 9^{\large 3027}\!\equiv\underbrace{(-1\!+\!10)^{\large 3027}\!\equiv -1 +\! \overbrace{3027}^{\large \color{#c00}{7}+10j}(\color{#c00}{10})}_{\rm Binomial\ Theorem\ }\equiv -1+\color{#c00}{70}\equiv 69$
A: You could use the Chinese remainder theorem and do the problem inside $\mathbb{Z}/25\mathbb{Z}$ and $\mathbb{Z}/4\mathbb{Z}$ separately. $27$ has order $20$ in $\mathbb{Z}/25\mathbb{Z}$ and order 2 in $\mathbb{Z}/4\mathbb{Z}$. We have $27^{2018} = 27^{18} = 19 \pmod{25}$ and $27^{2018} = 1 \pmod{4}$, so $27^{2018} = 69 \pmod{100}$. 
A: Because $\varphi(100)=40$ certainly the sequence of powers of $27$ will repeat after $40$ terms. A quick calculation shows that in fact it already repeats after $20$ terms. You could also use the Chinese remainder theorem to reduce the problem to computing $27^{2018}$ mod $25$ and mod $4$.
A: As $27=3^3,N=27^{2018}=3^{3\cdot2018}$
Using http://mathworld.wolfram.com/CarmichaelFunction.html, $\lambda(100)=20$
As $(3,100)=1,6054\equiv14\pmod{20},$
$N\equiv3^{14}\pmod{100}$
Now $3^{14}=9^7=-(1-10)^7\equiv-1+\binom71\cdot10^1\pmod{100}$
See also: Last 3 digits of $3^{999}$
Determine the last two digits of $3^{3^{100}}$
A: Working in the multiplicative group of integers modulo $100$, $(\Bbb Z /100 \Bbb Z)^\times$, you can discover all that is needed to solve the problem with elementary algebra.
If $n$ is an integer denote by $[n]$ its $\text{modulo-}100$ residue class in  ${\displaystyle \mathbb {Z} /100 \mathbb {Z}}$.
Exercise 1: The subset $\mathcal V = \bigr\{[20n+d]\mid n \in \Bbb Z \land d \in \{1,3,7,9\}\bigr\}$ of ${\displaystyle \mathbb {Z} /100 \mathbb {Z}}$ is actually a subset of $(\Bbb Z /100 \Bbb Z)^\times$, forming a subgroup with $20$ elements.
Since $[27] \in \mathcal V$, $\,27^{20} \equiv 1 \pmod{100}$.
To solve $27x \equiv 1 \pmod{100}$, observe that $x$ has the form $20n +3$. Multiplying,
$\tag 1 (20 + 7)(20n+3) \equiv 60 + 40n + 21 \equiv 81 + 40n \pmod{100}$
Examining $\text{(1)}$ we determine that $x \equiv 63 \pmod{100}$.
So
$\; 27^{2018} \equiv 27^{2020} \cdot 63^{2} \equiv 63 \cdot 63 \equiv 80 + 80 + 9 \equiv 69 \pmod{100}$.
