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?

Why are you taking powers of $27$ if you want a power of $2$? — J. W. Tanner, Mar 17 '19 at 20:34

score 4 · Accepted Answer · answered Mar 17 '19 at 20:57

4

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

answered Mar 17 '19 at 20:57

robjohn

345,667

Very clear and seems like a way faster method than I was using. Thanks. – Mathbeginner Mar 17 '19 at 21:57
@Mathbeginner The fastest way is probably via the Binomial Theorem, e.g. see my answer ($10$ seconds of very trivial mental arithmetic) – Bill Dubuque Mar 18 '19 at 02:34
@BillDubuque: works nicely because $3^2\equiv-1\pmod{10}$. It may not generalize to other bases than $3$. – robjohn Mar 18 '19 at 02:50
@Rob Indeed, but such exercises are usually designed to have a nice solution by hand. – Bill Dubuque Mar 18 '19 at 02:57
Is the downvote due to a problem or question with this answer? I think the square and multiply algorithm after reducing the exponent by the totient of the modulus is a decent and fairly algorithmic approach to this type of question. – robjohn Mar 19 '19 at 19:03
I don't see any problem. In case you might not know, there are at least a handful of users who don't like such questions, so perhaps they may use such votes to try to discourage folks who answer them (I occasionally get such votes too). – Bill Dubuque Mar 20 '19 at 19:48
@BillDubuque: I understand that, but those are usually for questions with no effort shown, and this question shows the attempt made. Because mine was the only answer downvoted, I was thinking it might be something other than just the question. – robjohn Mar 20 '19 at 20:26
Btw, to clarify, "such questions" above doesn't refer to general "low quality" questions but, rather, to questions on modular exponentiation. A few users have expressed opinions that most such questions should be closed as dupes of a generic "abstract dupe" thread. Further some think answering such questions is based purely on rep-hunting. As I've mentioned elsewhere, I disagree in general (though I might agree that some specific cases are dupes). – Bill Dubuque Mar 20 '19 at 22:35

TheSilverDoe · Answer 2 · 2019-03-17T20:52:06.797

2

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

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

edited Mar 17 '19 at 20:52

answered Mar 17 '19 at 20:47

TheSilverDoe

29,720

Do you know, without use of a calculator, that $27^{5}\equiv 7 \mod (100)$? – Mathbeginner Mar 17 '19 at 20:49
I think you meant $11^3$, not $33$ – J. W. Tanner Mar 17 '19 at 20:51
@Mathbeginner $27^5 = 27 \times (27^2)^2$. Now $27^2 = 729 = 29 \quad (\mathrm{mod} \text{ } 100)$. So $27^5 = 27 \times 29^2$ which you can compute by hand. – TheSilverDoe Mar 17 '19 at 20:53
@J.W.Tanner Yes, typo, thanks ! – TheSilverDoe Mar 17 '19 at 20:53

score 2 · Answer 3 · answered Mar 18 '19 at 02:31

2

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

answered Mar 18 '19 at 02:31

Bill Dubuque

272,048

+1: nice and different approach – robjohn Mar 18 '19 at 02:40

score 1 · Answer 4 · answered Mar 17 '19 at 20:44

1

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

answered Mar 17 '19 at 20:44

user555203

3,367

how do you see so quickly that $27$ has order $20$ in $\mathbb{Z}/25\mathbb{Z}$? – Mathbeginner Mar 17 '19 at 20:48
$2^{10} = 1024 = 24 = -1 \pmod{25}$. Or just compute the powers of $2$ mod $25$. Since $25$ is considerably smaller than $100$, it's much easier than what you were doing. – user555203 Mar 17 '19 at 20:49
I do see how $27$ has order $2$ in $\mathbb{Z}/4\mathbb{Z}$ since $729 \equiv 1 \mod (4)$ but I still don't see how what you're saying about $2^{10}$ is related to $27$ in $\mathbb{Z}/25\mathbb{Z}$... Sorry for the lack of insight. – Mathbeginner Mar 17 '19 at 21:00
You can reduce modulo $4$ (or $25$) before squaring (or doing any other arithmetic), and that's essentially why using the CRT is useful here. $27 = 3 \ne 1$ in $\mathbb{Z}/4\mathbb{Z}$, and $3^2 = 1$. That's exactly what it means to have order $2$. Similarly, in $\mathbb{Z}/25\mathbb{Z}$, we have $27 = 2$. And $2^{10} = 1024$. And $1024 = 24 = -1 \ne 1$ in $\mathbb{Z}/25\mathbb{Z}$. And $2^{20} = (2^{10})^2 = (-1)^2 = 1$. That's what it means to have order $20$. – user555203 Mar 17 '19 at 22:34

score 1 · Answer 5 · answered Mar 17 '19 at 20:45

1

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 $4$.

answered Mar 17 '19 at 20:45

Servaes

63,261
7
75
163

I'm not familiar with this $\varphi$, I do reckon you mean the totiënt function, right? Do you know by hearth that $\varphi(100)=40$? – Mathbeginner Mar 17 '19 at 20:53
2

Yes, totient function $\phi(100)=\phi(25)\phi(4)=(5\times4)(2\times1)$ – J. W. Tanner Mar 17 '19 at 21:23
Thanks, makes sense now! – Mathbeginner Mar 17 '19 at 21:58

score 1 · Answer 6 · answered Mar 18 '19 at 01:26

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 $3^{3^{100}}$

score 1 · Answer 7 · answered Oct 31 '20 at 03:11

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

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

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

7 Answers7

Linked