A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 3 years ago
How would you simplify it manually?
\[621^{1681} mod 2231\]
anonymous
 3 years ago
How would you simplify it manually? \[621^{1681} mod 2231\]

This Question is Closed

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Here's my attempt: (It's a bit long, so I separated it into three lines, but they should be in one line only) \[1681_{10} = 11010010001_{2}\] \[621^{1681} mod 2231 = 621^2mod2231)\times621mod2231)^2mod2231)^2\]\[\times 621mod2231)^2mod2231)^2mod2231)^2\]\[\times 621mod2231)^2mod2231)^2mod2231)^2mod2231)^2 \times 621mod2231\] Then, I got 1610 as the answer. Is there any faster way to do it manually?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0if im remembering the phi function correctly then if\[1681=\phi(2231)+k\] then we could reduce it to \[621^{k}\]

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.02231 = 23*97, prime factors since prime numbers are relatively prime to one another; and since the phi value of any prime number is one less than the prime; phi(2231) = phi(23)*phi(97) = 22*96 = 2112 but my memory aint cooperating with me to see this clearly :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0This reminds me of n = pq , where p and q are prime numbers. ϕ(n) = (p1)(q1) choose 1< e ≤ ϕ(n), such that gcd (e, ϕ(n)) =1 ed mod ϕ(n) =1 d is the multiplicative inverse of e. But.. the above is not related to the question, I guess?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0the RSA encryption uses the thrm yes; and i was just wondering if the thrm was applicable to simplify this. If we could simplify it to:\[\Large 621^{k\phi(2231)+m}\]then we might be able to play about with it ... just a thought :)

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.0Looking at this, and assuming we could factor these numbers, I might suggest the chinese remainder theorem.\[621^{1681}\pmod{2231}\implies621^{1681}\pmod{23*97}\]Since \(23621\), we have that \[621^{1681}\equiv0\pmod{23},\]and we can calculate (it's fairly tedious, but possible by hand) that \[621^{1681}\equiv 58\pmod{97}.\]So we want some \(x\) such that \[x\equiv 23s\pmod{23}\]\[x\equiv58+97t\pmod{97}.\]Thus\[23s\equiv58\pmod{97}\]Another (tedious, but possible to do by hand) calculation shows that this means that \(s\equiv70\pmod{97}\). Then we plug it back into the original equation to get \(x=23(70)=1610\).

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.0Give me a hoot if you want an explanation of how to solve those two steps by hand.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Will revisit this question again next week, if you don't mind...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I can work out the line \(621^{1681}≡58 \pmod{97}\) using the method in my first post. Is there any alternative way to do it? For the rest of your first comment, since I haven't learnt the Chinese Remainder Theorem, I don't know why and how it works. I'm sorry that I need some more time to do a bit selfstudy before coming back to this question again.

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.0Well, since 97 is prime, and \(97\nmid 621\), we know that \(621^{96}\equiv1\pmod{97}\). So we find \(1681\equiv 49\pmod{96}\), and hence, \[621^{1681}\equiv39^{49}\pmod{97}.\]Then (I think this is fundamentally the same thing you did) you say \(49=2^5+2^4+2^0\). Thus, this is the same as\[39^{2^5}39^{2^4}39\pmod{97}.\]Then by repeatedly squaring, and reducing mod 97, you can find this number. Again, I think this is the same, in principle, of what you did.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.