Solution to Modular Congruence: $3^{47}\mod23$

Question

I am trying to solve the following modular congruence but I am not sure how to proceed as I have not fully grasp the concept of it.

$$\large3^{47} \mod 23$$

I would like to know how to proceed in solving this while understanding the process in doing so.

Somewhat, but I don't fully understand it. Would you mind explaining? — Omari Celestine, Oct 31 '17 at 00:48
A special case of the theorem says that, when $p$ is prime, $a^{p-1} \equiv 1 \pmod{p}$ for any nonzero $a \in \mathbb{Z}$. In this problem, a good first step would be recognizing that $3^{47} = 3^{22 + 22 + 3} = 3^{22}3^{22}3^3$ and applying that fact. — Kaj Hansen, Oct 31 '17 at 00:50
Do you think you can post an answer demonstrating the full process? — Omari Celestine, Oct 31 '17 at 00:55
@OmariCelestine It would be a better idea to try the hint first, and then ask again if you don't understand. This will help you better understand the process, not the answer. — Toby Mak, Oct 31 '17 at 01:33

score 3 · Answer 1 · answered Oct 31 '17 at 01:16

3

Another way of stating Fermat's Little Theorem is $a^p\equiv a$ mod $p$ for any $a$ if $p$ is prime. Thus we have

$$3^{47}=3^{23+23+1}=3^{23}\cdot3^{23}\cdot3\equiv3\cdot3\cdot3=27\equiv4\mod23$$

answered Oct 31 '17 at 01:16

Barry Cipra

Bernard · Accepted Answer · 2017-10-31T01:00:35.690

2

Hint:

Use lil' Fermat: as $23$ is prime and $3$ is not divisible by $23$, $$3^{47}\equiv3^{47\bmod 22}\mod23.$$

Some details:

$47\bmod 22=3$, so $\;3^{47}\equiv 3^3=27\equiv 4\mod23$.

edited Oct 31 '17 at 01:00

answered Oct 31 '17 at 00:46

Bernard

And that would be the solution? – Omari Celestine Oct 31 '17 at 00:50
You have to end the computation, naturally. – Bernard Oct 31 '17 at 00:51
could you please demonstrate? As I mentioned, I have not grasped the concept of modular congruence as yet. – Omari Celestine Oct 31 '17 at 00:53
I've added the computation. All you have to know is that $a\equiv b\mod 23$ means $a$ and $b$ have the same remainder in the division by $23$, and $a\bmod23$ denotes this remainder. Is that clearer? – Bernard Oct 31 '17 at 01:00
@Randall: Right. I was thinking of $\varphi(23)$. Fixed. Thanks for pointing it! – Bernard Oct 31 '17 at 01:01
What does φ(23) mean? Also, where did the $3^{3}$ come from? – Omari Celestine Oct 31 '17 at 01:15
It is Euler's totient function which gives the number of units modulo $n$. For a prime number $p$, its value is $p-1$. The exponent of $3$ is the remainder when you divide $47$ by $22$. – Bernard Oct 31 '17 at 01:18

2 Answers2