Find the remainder in the division of $3^{385}$ by $400$.

Question

What I thought:
I'm looking for the smallest $r$ such that $3^{385} \equiv r \pmod {400}$. Which is equivalent to the system $\begin{cases} 3^{385} \equiv r \pmod {2^4} \\ 3^{385} \equiv r \pmod {5^2}\end{cases}$. Obviously, $3^{385} \equiv 1 \pmod {2}$. Using Fermat's Little Theorem, $3^{385} \equiv 3^1 \pmod {5}$ since $385 \equiv 1 \pmod 4$. However, I'm not sure how to extend this to the original system. I've figured that $r \mod 2^4 \in \{1,3,5,7,9,11,13,15\}$ and $r \mod 5^2\in \{3,8,13,18,23\}$. But how do I know which?

Thanks for the help!!

Edit: Look, I should have clarified this earlier: there are too many ways to solve this, and I in fact know many of them. What I need is a follow up on what I have done. How do I solve the system of congruences when the factors of the original modulus are powers?

Try computing $3^{385}$ modulo 16 and 25 and using the Chinese Remainder Theorem. You can use the generalised version of Fermat's Little Theorem with the Euler Totient Function to compute these values. — Mark Saving, Jul 26 '21 at 19:20
Does this answer your question? Modular exponentiation by hand ($a^b\bmod c$) — Arthur, Jul 26 '21 at 19:36
The Carmichael function of $400$ is $20$, so $3^{385}\equiv 3^{380}3^5\equiv 3^5\equiv243\bmod400$ — J. W. Tanner, Jul 26 '21 at 19:38
$3^{385}=3\cdot(10-1)^{192}\equiv3\cdot(1-10\times192)=-5757\equiv6000-5757\bmod400$ — J. W. Tanner, Jul 26 '21 at 20:01
@MarkSaving Computing $3^385$ modulo 16 and 25 is the one thing I can't do, otherwise I'd very much be able to use the CRT. However, I don't know about the Euler Totient Function, but I know for a fact that I'm not supposed to use that. — BrKo14, Jul 26 '21 at 20:14
$3^{4}=81\equiv1\bmod16$, so $3^{385}\equiv3^{384}3\equiv3\bmod16$ — J. W. Tanner, Jul 26 '21 at 20:16

score 1 · Answer 1 · 2022-04-04T05:55:51.310

1

$$3^{385} = 243^{77}$$ $$\text{We must compute} \; 243^{76}\cdot 243 \; \text{mod}\,(400)$$ $$\Rightarrow 243^{(4\cdot19)} \cdot 243 \; \text{mod}\,(400)$$ $$= (243^4)^{19}\; \text{mod}\,(400) \cdot 243 \; \text{mod}\,(400)$$ $$= (1)^{19} \; \text{mod}\,(400) \cdot 243 \; \text{mod}\,(400)$$ $$= 243 \; \text{mod}\,(400)$$ $$\boxed{\therefore 3^{385} \equiv 243 \; \text{mod}\,(400)}$$

edited Apr 04 '22 at 05:55

answered Jul 26 '21 at 19:32

2

This answer looks like you pulled $243$ out of a hat and found that it works. Please edit your answer and explain why you broke down $3^{385}$ as $(3^5)^{77}$ and not $3\cdot(3^4)^{96}$ or $(3^7)^{55}$ – Moko19 Jul 26 '21 at 19:40
Sure, I can break the number to somthing small enough that I can calculate. But this is not the answer I'm looking for. – BrKo14 Jul 26 '21 at 20:11

score 1 · Answer 2 · answered Jul 26 '21 at 19:50

1

Reduced totient for 400 is 20.

So $3^{20} \equiv 1\mod400$ per Carmichael's Theorem.

Hence $3^{380} \equiv 1\mod400$, which implies $3^{385} \equiv 3^5 = 243\mod400$.

answered Jul 26 '21 at 19:50

Shang Zhang

496
2
5

score 0 · Answer 3 · answered Jul 26 '21 at 19:37

0

Hint:

$$3^{385}\equiv 3^{385\bmod\operatorname{order}(3\bmod 400)}\mod 400.$$ Further, the order of $3$ modulo $400$ is the l.c.m. of its orders $\bmod 25$ and $\bmod 16$.

answered Jul 26 '21 at 19:37

Bernard

175,478

score 0 · Answer 4 · answered Jul 26 '21 at 19:46

0

You are on the right track. But when a prime power ($2^4$ or $5^2$ in this instance) divides the modulus, you have to compute the answer modulo each prime power.

So you have to compute $3^{385} \bmod 16$ and $3^{385}\bmod 25$. We have $\varphi(16)=8$ and $\varphi(25)=20$; can you take it from there?

answered Jul 26 '21 at 19:46

TonyK

64,559

Excuse my ignorance, what's the $\varphi$ symbol here? – BrKo14 Jul 26 '21 at 20:09
1

@BrKo14 Euler's totient function. – Theo Bendit Jul 26 '21 at 20:14

Find the remainder in the division of $3^{385}$ by $400$.

4 Answers4