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ParthKohli Group Title

A common technique for finding modulo \(10\) of any huge number in the form \(a^b\)?

  • one year ago
  • one year ago

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  1. ParthKohli Group Title
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    It seems that you have to go through a lot of manipulations... a lot for doing that. Although I was thinking of a technique they use to figure all that out.

    • one year ago
  2. ParthKohli Group Title
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    OK, I'm gonna change the topic to something else:\[39^{39} \equiv n\pmod{100}\]

    • one year ago
  3. ParthKohli Group Title
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    \(n\) has to be a two-digit number. (I actually have to find the last two digits)

    • one year ago
  4. ParthKohli Group Title
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    Something with a pattern?

    • one year ago
  5. shubhamsrg Group Title
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    39 = -61 mod100 =>39^38 = 61^38 mod100 =>39^39 = 39*61^38 mod100 SInce 61^n always ends in 1, that multiplied by 39, will give last digit as 9, hmm, leme see for the second last.

    • one year ago
  6. terenzreignz Group Title
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    Better yet, 39^5 = -1 mod 100 Use that: 39^39 = (39^5)^7 * 39^4 = (-1)^7 * 39^4 mod 100

    • one year ago
  7. shubhamsrg Group Title
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    Seems legit. But on paper, who's gonna go till 39^5 ! :O

    • one year ago
  8. ParthKohli Group Title
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    Very nice, can you tell me *how* you came up with that solution? What do you initially think?

    • one year ago
  9. terenzreignz Group Title
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    square 39, and then only take the last two digits...

    • one year ago
  10. terenzreignz Group Title
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    then multiply 39 to it again.

    • one year ago
  11. terenzreignz Group Title
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    and only take the last two digits, again...

    • one year ago
  12. ParthKohli Group Title
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    Ahem, that was beautiful. But is there a common way to figure out these solutions?

    • one year ago
  13. shubhamsrg Group Title
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    Ohh, I see.

    • one year ago
  14. terenzreignz Group Title
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    The way I think of things, and it's a common trend that the way I do it is highly inefficient (LOL) But when faced with a problem b^p = n (mod k) I try to see if there's an exponent to which b can be raised so that it is 1 (or -1) mod k first.

    • one year ago
  15. terenzreignz Group Title
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    That "last two digits" trick only works because you were conveniently working under mod 100... XD

    • one year ago
  16. ParthKohli Group Title
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    Nice... can you give me another example?

    • one year ago
  17. terenzreignz Group Title
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    How about 6^5000 = n (mod 215)

    • one year ago
  18. terenzreignz Group Title
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    (sneakily thinking of simple problems) :>

    • one year ago
  19. ParthKohli Group Title
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    Arhmegurd.\[\left(6^{5}\right)^{1000} \equiv (6^2)^{1000} \equiv n \pmod{215}\]

    • one year ago
  20. ParthKohli Group Title
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    \[\left(6^5\right)^{400} \equiv (6^2)^{400} \equiv n\pmod{215}\]

    • one year ago
  21. ParthKohli Group Title
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    \[\left(6^5\right)^{120} \equiv 6^{240} \equiv n \pmod{215}\]

    • one year ago
  22. terenzreignz Group Title
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    I'm not following o.O You may be using a method I'm not familiar with, though... if you are, do continue :)

    • one year ago
  23. ParthKohli Group Title
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    \[\left(6^{5}\right)^{48} \equiv 6^{96} \equiv n \pmod{215}\]

    • one year ago
  24. ParthKohli Group Title
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    Hmm... I am just using \(6^2 \equiv 6^5 \pmod{215}\)

    • one year ago
  25. ParthKohli Group Title
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    What should I do here?

    • one year ago
  26. terenzreignz Group Title
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    Hang on...

    • one year ago
  27. terenzreignz Group Title
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    I was rather hoping you'd go for \[6^3 = 216 = -1(mod \ 215)\]

    • one year ago
  28. ParthKohli Group Title
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    Oh...

    • one year ago
  29. terenzreignz Group Title
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    Wait, typo -.-

    • one year ago
  30. ParthKohli Group Title
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    Yes, isn't that 1 (mod 215)

    • one year ago
  31. terenzreignz Group Title
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    \[\large 6^3 = 216 = 1(\mod \ 215 \ \ \ \ )\] Much better

    • one year ago
  32. terenzreignz Group Title
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    And work from there...

    • one year ago
  33. terenzreignz Group Title
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    \[\large 6^{5000} = 6^{3(1666)+2}\]

    • one year ago
  34. ParthKohli Group Title
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    \[6^{5000}= 6^{4998} \times 36 \equiv \]

    • one year ago
  35. terenzreignz Group Title
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    \[1 \times 36 (\mod 215 \ \ \ \ \ )\]

    • one year ago
  36. ParthKohli Group Title
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    \[\equiv 36 \pmod{215}\]

    • one year ago
  37. ParthKohli Group Title
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    Argh, you type fast. =3

    • one year ago
  38. terenzreignz Group Title
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    Awesomeness

    • one year ago
  39. ParthKohli Group Title
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    Another n00bish question coming up. Thanks man.

    • one year ago
  40. terenzreignz Group Title
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    If you have time, look up Fermat's Little Theorem, might come in handy in a pinch.

    • one year ago
  41. ParthKohli Group Title
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    Yesh, that one is cool:\[p^{n - 1} \equiv 1 \pmod{n} \]I'd try to see if that applies in the future.

    • one year ago
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