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ParthKohli

  • 2 years ago

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

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  1. ParthKohli
    • 2 years ago
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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.

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

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

  4. ParthKohli
    • 2 years ago
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    Something with a pattern?

  5. shubhamsrg
    • 2 years ago
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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.

  6. terenzreignz
    • 2 years ago
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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

  7. shubhamsrg
    • 2 years ago
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    Seems legit. But on paper, who's gonna go till 39^5 ! :O

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

  9. terenzreignz
    • 2 years ago
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    square 39, and then only take the last two digits...

  10. terenzreignz
    • 2 years ago
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    then multiply 39 to it again.

  11. terenzreignz
    • 2 years ago
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    and only take the last two digits, again...

  12. ParthKohli
    • 2 years ago
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    Ahem, that was beautiful. But is there a common way to figure out these solutions?

  13. shubhamsrg
    • 2 years ago
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    Ohh, I see.

  14. terenzreignz
    • 2 years ago
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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.

  15. terenzreignz
    • 2 years ago
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    That "last two digits" trick only works because you were conveniently working under mod 100... XD

  16. ParthKohli
    • 2 years ago
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    Nice... can you give me another example?

  17. terenzreignz
    • 2 years ago
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    How about 6^5000 = n (mod 215)

  18. terenzreignz
    • 2 years ago
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    (sneakily thinking of simple problems) :>

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

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

  21. ParthKohli
    • 2 years ago
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    \[\left(6^5\right)^{120} \equiv 6^{240} \equiv n \pmod{215}\]

  22. terenzreignz
    • 2 years ago
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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 :)

  23. ParthKohli
    • 2 years ago
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    \[\left(6^{5}\right)^{48} \equiv 6^{96} \equiv n \pmod{215}\]

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

  25. ParthKohli
    • 2 years ago
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    What should I do here?

  26. terenzreignz
    • 2 years ago
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    Hang on...

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

  28. ParthKohli
    • 2 years ago
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    Oh...

  29. terenzreignz
    • 2 years ago
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    Wait, typo -.-

  30. ParthKohli
    • 2 years ago
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    Yes, isn't that 1 (mod 215)

  31. terenzreignz
    • 2 years ago
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    \[\large 6^3 = 216 = 1(\mod \ 215 \ \ \ \ )\] Much better

  32. terenzreignz
    • 2 years ago
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    And work from there...

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

  34. ParthKohli
    • 2 years ago
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    \[6^{5000}= 6^{4998} \times 36 \equiv \]

  35. terenzreignz
    • 2 years ago
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    \[1 \times 36 (\mod 215 \ \ \ \ \ )\]

  36. ParthKohli
    • 2 years ago
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    \[\equiv 36 \pmod{215}\]

  37. ParthKohli
    • 2 years ago
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    Argh, you type fast. =3

  38. terenzreignz
    • 2 years ago
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    Awesomeness

  39. ParthKohli
    • 2 years ago
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    Another n00bish question coming up. Thanks man.

  40. terenzreignz
    • 2 years ago
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    If you have time, look up Fermat's Little Theorem, might come in handy in a pinch.

  41. ParthKohli
    • 2 years ago
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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.

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