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KingGeorge

[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: \(713=23\cdot31\).

  • one year ago
  • one year ago

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  1. KingGeorge
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    Since 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.

    • one year ago
  2. KingGeorge
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    Hint 2: Use the Chinese Remainder Theorem. You will find it isn't as bad as it usually is.

    • one year ago
  3. zzr0ck3r
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    what is mod? what class do we learn about this?

    • one year ago
  4. KingGeorge
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    It'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.

    • one year ago
  5. KingGeorge
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    It's also called "Clock arithmetic" sometimes http://en.wikipedia.org/wiki/Modular_arithmetic

    • one year ago
  6. zzr0ck3r
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    ahhh ok we used this in computer science

    • one year ago
  7. zzr0ck3r
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    so 8mod(3) = 2?

    • one year ago
  8. KingGeorge
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    Right.

    • one year ago
  9. zzr0ck3r
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    hmm

    • one year ago
  10. zzr0ck3r
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    one more hint, i dont see how the 23* 31 helps unless we could break up the top. into some multiple of 23 or 31

    • one year ago
  11. KingGeorge
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    You 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.

    • one year ago
  12. zzr0ck3r
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    thats the fun of it:) I got to eat but will be back in a bit to try more

    • one year ago
  13. KingGeorge
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    Extra 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.

    • one year ago
  14. Ishaan94
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    \[89^{307}\pmod{23}=15\]\[89^{307}\pmod{31} = 15\]15?

    • one year ago
  15. KingGeorge
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    Right 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.

    • one year ago
  16. Ishaan94
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    Thank you for helping me through.

    • one year ago
  17. KingGeorge
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    How much were you able to do by hand?

    • one year ago
  18. Ishaan94
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    I used calculator for 13*12*20 (mod 23) and for verifying my answers.

    • one year ago
  19. KingGeorge
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    Excellent job! However, there was an alternative, if slightly roundabout way of finding \(89^{307}\pmod{23}\) without using successive squaring.

    • one year ago
  20. Ishaan94
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    oh hmm can you show me? or maybe a hint?

    • one year ago
  21. KingGeorge
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    Basically, 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\[20x\equiv1\pmod{23}\]

    • one year ago
  22. KingGeorge
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    Got to go eat dinner now. Feel free to ask more questions on this and I'll answer them when I get back.

    • one year ago
  23. Ishaan94
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    I like it, it's much better and simpler than mine. Thank you.

    • one year ago
  24. KingGeorge
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    I love it when I can use it. It was a little trick I learned in my cryptography class.

    • one year ago
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