Computing last two digits of $27^{2018}$

Question

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?

Answer 1

$\phi(100)=40$, so we can reduce the exponent mod $40$: $$ 27^{2018}\equiv27^{18}\pmod{100} $$ Then we can square and multiply: $$ \begin{align} 27^2&\equiv29\\ 27^4&\equiv41\\ 27^8&\equiv81\\ 27^9&\equiv87\\ 27^{18}&\equiv69\pmod{100} \end{align} $$

Answer 2

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 $27^{2018} = 69 \pmod{100}$.

Answer 3

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 $4$.

Answer 4

$27^{5} = 7 \quad (\mathrm{mod} \text{ } 100)$.

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 $$27^{2018} = 69 \quad (\mathrm{mod} \text{ } 100)$$

Answer 5

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 $3^{3^{100}}$

Answer 6

$\bmod 100\!:\,\ 3^{\large 3\cdot 2018}\!\equiv 9^{\large 3027}\!\equiv\underbrace{(-1\!+\!10)^{\large 3027}\!\equiv -1 +\! \overbrace{3027}^{\large \color{#c00}{7}+10j}(\color{#c00}{10})}_{\rm Binomial\ Theorem\ }\equiv -1+\color{#c00}{70}\equiv 69$

Answer 7

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

$\; 27^{2018} \equiv 27^{2020} \cdot 63^{2} \equiv 63 \cdot 63 \equiv 80 + 80 + 9 \equiv 69 \pmod{100}$.