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goformit100

  • 2 years ago

Find the remainder when 2^1990 is divided by 1990.

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  1. goformit100
    • 2 years ago
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    @mayankdevnani

  2. ParthKohli
    • 2 years ago
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    Mod arithmetic :') @terenzreignz

  3. terenzreignz
    • 2 years ago
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    Why me? :/

  4. ParthKohli
    • 2 years ago
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    Because you.

  5. goformit100
    • 2 years ago
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    In this question How to square to so much power ?

  6. terenzreignz
    • 2 years ago
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    Might have to resort to totients..... @ParthKohli ?

  7. goformit100
    • 2 years ago
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    Sir if i use the exponent rule will it work here ?

  8. ParthKohli
    • 2 years ago
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    Ah! Euler's Theorem!

  9. satellite73
    • 2 years ago
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    factor 1990 first

  10. terenzreignz
    • 2 years ago
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    Time to doodle... \[\large 2^{1990}=4^{995}\]

  11. goformit100
    • 2 years ago
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    How to factor it ?

  12. satellite73
    • 2 years ago
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    how to factor 1990?

  13. terenzreignz
    • 2 years ago
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    What is 4^5? \[\Large = 1024^{199}\]

  14. goformit100
    • 2 years ago
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    Yes @satellite73

  15. satellite73
    • 2 years ago
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    try \(2\times 5\times 199\)

  16. mayankdevnani
    • 2 years ago
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    1990 = 10*199 = 2* 5* 199

  17. terenzreignz
    • 2 years ago
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    \[\large 1024^{199}=1024\cdot 1024^{198}=1024\cdot 2048^{99}\]

  18. goformit100
    • 2 years ago
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    try \(2\times 5\times 199\) means ?

  19. terenzreignz
    • 2 years ago
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    Now let's start working some "mod magic" and reduce the bases at mod 1990 \[\Large =_{(mod \ 1990)} \ \ 1024\cdot 58^{99}\]

  20. mayankdevnani
    • 2 years ago
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    |dw:1367281405845:dw|

  21. mayankdevnani
    • 2 years ago
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    ok @goformit100

  22. goformit100
    • 2 years ago
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    ok So LCM can be take as a good way of factorizing numbers ok ?

  23. terenzreignz
    • 2 years ago
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    \[\Large \equiv 1024\cdot 58 \cdot 58^{98}\]

  24. mayankdevnani
    • 2 years ago
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    yaaa @goformit100

  25. mayankdevnani
    • 2 years ago
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    it is the best way!!!

  26. terenzreignz
    • 2 years ago
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    \[\Large \equiv 1682 \cdot 1374^{49}\] gah... possibly inefficient...

  27. goformit100
    • 2 years ago
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    ok

  28. terenzreignz
    • 2 years ago
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    \[\Large\equiv 678\cdot (-616)^{48}\]

  29. terenzreignz
    • 2 years ago
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    This is daunting.

  30. satellite73
    • 2 years ago
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    \(1990=2\times 5\times 199\) and \(2^2\equiv 4(5) \) also \(2^{198}\equiv 1(199)\) by fermat

  31. goformit100
    • 2 years ago
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    Mods Sir(s) I have to make you know that the Equation you are posting have not opened yet

  32. satellite73
    • 2 years ago
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    actually this is kind of a pain isn't it

  33. amistre64
    • 2 years ago
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    if you are seeing "math processing error", try refreshing

  34. goformit100
    • 2 years ago
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    What to do ?

  35. terenzreignz
    • 2 years ago
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    \[\Large \equiv 678\cdot (1356)^{24}\]

  36. goformit100
    • 2 years ago
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    Yo It's done...Mods you'll great REFRESHING WORKED. No SEE

  37. goformit100
    • 2 years ago
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    Now I can see

  38. terenzreignz
    • 2 years ago
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    \[\Large \equiv 678 \cdot (-634)^{24}\equiv678\cdot (634)^{24}\]

  39. goformit100
    • 2 years ago
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    2^1990 is divided by 1990 what to actually for this ?

  40. amistre64
    • 2 years ago
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    2^1990 2^11 = 2048 = 38 mod 1990 2^(11(180)+10) is what i had in mond :)

  41. terenzreignz
    • 2 years ago
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    I lack creativity, guys :)

  42. goformit100
    • 2 years ago
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    @mikaela19900630 you may too se the question I have posted now.

  43. goformit100
    • 2 years ago
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    Thank you all of you. I can do these type of question Now :)

  44. goformit100
    • 2 years ago
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    *from Now

  45. terenzreignz
    • 2 years ago
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    \[\Large \equiv 678\cdot (-24)^{24}\equiv 678\cdot 24^{24}\]

  46. terenzreignz
    • 2 years ago
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    cr*p... sorry \[\Large \equiv 678\cdot (-24)^{\color{red}{12}}\equiv 678\cdot 24^{\color{red}{12}}\]

  47. goformit100
    • 2 years ago
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    Thank You Very Much.

  48. goformit100
    • 2 years ago
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    COME TO http://openstudy.com/study#/updates/517e64aae4b05fd789937825

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