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 2 years ago
[SOLVED] Number Theory Challenge (not actually that hard)
Find \(89^{307}\pmod{713}\) without using any kind of electronic aid.
Please at least attempt it before resorting to Wolfram.
Hint: \(713=23\cdot31\).
 2 years ago
[SOLVED] Number Theory Challenge (not actually that hard) Find \(89^{307}\pmod{713}\) without using any kind of electronic aid. Please at least attempt it before resorting to Wolfram. Hint: \(713=23\cdot31\).

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KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.2Since I ran out of space on the question: I only decided to ask this because when I was first doing this problem, I quickly discovered I didn't need an electronic aid.

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.2Hint 2: Use the Chinese Remainder Theorem. You will find it isn't as bad as it usually is.

zzr0ck3r
 2 years ago
Best ResponseYou've already chosen the best response.0what is mod? what class do we learn about this?

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.2It's a number theory function. So \(a\pmod{n}\) returns the remainder of some number a when divided by n. There are a bunch of interesting applications for this, and when \(n\) is prime, you get a bunch of awesome properties.

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.2It's also called "Clock arithmetic" sometimes http://en.wikipedia.org/wiki/Modular_arithmetic

zzr0ck3r
 2 years ago
Best ResponseYou've already chosen the best response.0ahhh ok we used this in computer science

zzr0ck3r
 2 years ago
Best ResponseYou've already chosen the best response.0one more hint, i dont see how the 23* 31 helps unless we could break up the top. into some multiple of 23 or 31

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.2You need to look at \(89^{307}\pmod{23}\) and \(89^{307}\pmod{31}\) independently. Since these are prime, a bunch of nice properties apply, and you can simplify each to a number between 0 and 23/31. Once you have that, apply Chinese Remainder Theorem. I've got to say though, if this is the first time you've ever really used modular arithmetic, this is still a hard problem.

zzr0ck3r
 2 years ago
Best ResponseYou've already chosen the best response.0thats the fun of it:) I got to eat but will be back in a bit to try more

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.2Extra hints just for you. You need to be somewhat familiar with Fermat's Little Theorem. You also need to know how to find an inverse mod p using the Extended Euclidean Algorithm, and use successive squaring. These are all fairly basic number theory algorithms, and with a little practice, you should be able to get them.

Ishaan94
 2 years ago
Best ResponseYou've already chosen the best response.1\[89^{307}\pmod{23}=15\]\[89^{307}\pmod{31} = 15\]15?

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.2Right on. Technically it requires an application of CRT at the end (which is not easy to do by hand), but since both are congruent to 15, you can skip that step.

Ishaan94
 2 years ago
Best ResponseYou've already chosen the best response.1Thank you for helping me through.

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.2How much were you able to do by hand?

Ishaan94
 2 years ago
Best ResponseYou've already chosen the best response.1I used calculator for 13*12*20 (mod 23) and for verifying my answers.

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.2Excellent job! However, there was an alternative, if slightly roundabout way of finding \(89^{307}\pmod{23}\) without using successive squaring.

Ishaan94
 2 years ago
Best ResponseYou've already chosen the best response.1oh hmm can you show me? or maybe a hint?

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.2Basically, notice the following.\[89^{307}\equiv 20^{21}\pmod{23}\]By Fermat's Little Theorem, you have that \[20^{22}\equiv1\pmod{23}\]so\[20^{22}\equiv20^{21}\cdot20\equiv1\pmod{23}\]Hence, \[20^{21}\equiv20^{1}\pmod{23}\]So you just need to solve the congruence\[20x\equiv1\pmod{23}\]

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.2Got to go eat dinner now. Feel free to ask more questions on this and I'll answer them when I get back.

Ishaan94
 2 years ago
Best ResponseYou've already chosen the best response.1I like it, it's much better and simpler than mine. Thank you.

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.2I love it when I can use it. It was a little trick I learned in my cryptography class.
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