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myininaya

  • 5 years ago

Polpak's Challenge Question Round 1 Let n be a positive integer. If we divide n by 3, the remainder is 2 If we divide n by 5, the remainder is 1 what is n?

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  1. myininaya
    • 5 years ago
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    or i mean what is the smaller possible value for n?

  2. myininaya
    • 5 years ago
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    smallest*

  3. anonymous
    • 5 years ago
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    Looks like the Chinese remainder theorem.

  4. myininaya
    • 5 years ago
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    yay! :)

  5. myininaya
    • 5 years ago
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    lol very good yes i used euclidean's algorithm which is mixed up with that Chinese stuff

  6. myininaya
    • 5 years ago
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    i really like this problem

  7. anonymous
    • 5 years ago
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    yeah I did something similar a few weeks ago to implement by hand RSA encryption

  8. myininaya
    • 5 years ago
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    omg polpak i love the RSA

  9. myininaya
    • 5 years ago
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    i want to be an expert at number theory

  10. anonymous
    • 5 years ago
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    pretty cool. You can work for nsa ;)

  11. myininaya
    • 5 years ago
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    my instructor may me look at the RSA because I said I was interested in number theory and i was kindof aftraid because i thought i was going to have to know some computer stuff, but all it really was was number theory i looked at the DSA which is fun too but not as fun as the RSA

  12. myininaya
    • 5 years ago
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    I also looked at the secure hash algorithm 1 which is really just discrete math but still fun :)

  13. myininaya
    • 5 years ago
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    i'm one of those people who get's the joke: There are 10 kinds of people in this world; those who know binary and those who don't

  14. anonymous
    • 5 years ago
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    all algorithms are just descrete math stuff ;p

  15. myininaya
    • 5 years ago
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    yes thats true

  16. myininaya
    • 5 years ago
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    what education level are you if you don't mind me asking?

  17. anonymous
    • 5 years ago
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    About to transfer to 4 year university.

  18. anonymous
    • 5 years ago
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    just finished the local CC.

  19. myininaya
    • 5 years ago
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    what?! no way

  20. myininaya
    • 5 years ago
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    how are you super smart?

  21. anonymous
    • 5 years ago
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    I'm old. Some smarts come with age. I also study a lot (comes with being old)

  22. myininaya
    • 5 years ago
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    i think you are really cool :)

  23. anonymous
    • 5 years ago
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    Well thanks. I'm 33 and started college a few years ago when my son was born.

  24. myininaya
    • 5 years ago
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    are you going to be a math person? i mean you really already are, but do something with math?

  25. myininaya
    • 5 years ago
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    math instructor? engineer? ...

  26. anonymous
    • 5 years ago
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    Shooting for a masters in computer science, though I'm thinking I'll probably do math too. Maybe two masters! =)

  27. myininaya
    • 5 years ago
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    you can do it you have the knowledge already for math

  28. myininaya
    • 5 years ago
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    you probably know how to solve differential equations already i think you done one here before dont really remember

  29. myininaya
    • 5 years ago
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    or some differential equations

  30. anonymous
    • 5 years ago
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    I've done 1 class in diff eq's and 1 class in linear algebra thus far

  31. myininaya
    • 5 years ago
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    are you american?

  32. myininaya
    • 5 years ago
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    americans are lazy. you can't be american because you did that self study stuff

  33. anonymous
    • 5 years ago
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    lol, yes I'm american. and yes, I'm lazy. most smart people are ;)

  34. myininaya
    • 5 years ago
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    lol

  35. anonymous
    • 5 years ago
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    I study cause it makes things easier

  36. anonymous
    • 5 years ago
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    it's a lot more work to not know how to do the homework and struggle through it for hours and hours

  37. myininaya
    • 5 years ago
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    yes i agree

  38. myininaya
    • 5 years ago
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    i do have a another challenge question for you if you are willing to accept?

  39. anonymous
    • 5 years ago
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    oh sure, though if it takes as long as euclid's algorithm I may have to hold off a while ;)

  40. myininaya
    • 5 years ago
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    \[\int\limits_{}^{} e^{\sqrt{3x+9}} dx\]

  41. myininaya
    • 5 years ago
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    evaluate the above is the question

  42. myininaya
    • 5 years ago
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    or the problem

  43. myininaya
    • 5 years ago
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    let me know if you want a hint k?

  44. anonymous
    • 5 years ago
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    sorry, had a crisis..

  45. myininaya
    • 5 years ago
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    its cool

  46. anonymous
    • 5 years ago
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    looks like just u-sub then by parts..

  47. anonymous
    • 5 years ago
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    \[u = \sqrt{3x + 9} \implies du = \frac{3}{u}dx \implies dx = \frac{u}{3}du\] \[\implies \int e^{\sqrt{3x+9}} dx = \frac{1}{3}\int ue^udu\] Then go with by-parts.

  48. anonymous
    • 5 years ago
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    not nearly as bad as some of the triple integrals I had last year in calc 3.

  49. anonymous
    • 5 years ago
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    those things took pages =(

  50. anonymous
    • 5 years ago
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    Whoops, should be 2u/3 I think actually

  51. myininaya
    • 5 years ago
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    wait i think you can teach me something one sec

  52. myininaya
    • 5 years ago
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    1 Attachment
  53. myininaya
    • 5 years ago
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    then integration by parts blah blah

  54. anonymous
    • 5 years ago
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    some mistakes there

  55. myininaya
    • 5 years ago
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    the only mistake i see is du=3/[2(3x+9)] dx but i really didn't use this simplification

  56. anonymous
    • 5 years ago
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    It should be \[\frac{2}{3}\int ue^udu\]

  57. anonymous
    • 5 years ago
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    What you have on the right side, second line is correct. but recall that sqrt(3x + 9) is u, so you just have u in the denominator

  58. myininaya
    • 5 years ago
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    oh darn it lol

  59. myininaya
    • 5 years ago
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    you way is much more easier instead of multiply by 1 you just freaking squared both sides of u=sqrt(3x+9) and got u^2=3x+9 so 2 du=3dx so dx=2du/3 you are freaking brillant lol not really i just didn't think of that i mean you are smart though

  60. anonymous
    • 5 years ago
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    I don't square nothin

  61. anonymous
    • 5 years ago
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    I think you misunderstand what I did

  62. myininaya
    • 5 years ago
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    me too

  63. myininaya
    • 5 years ago
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    but that squaring does work also

  64. anonymous
    • 5 years ago
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    my way is easier ;p

  65. anonymous
    • 5 years ago
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    Ok, Let \(u = \sqrt{3x + 9}\) \[\implies du = (\frac{1}{2\sqrt{3x+9}} * 3 )dx\] \[ = \frac{3}{2u}dx\] \[\implies dx = \frac{2u}{3}du\]

  66. anonymous
    • 5 years ago
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    Then go back to the integral and sub that for dx, and the exponent of e just becomes u also.

  67. myininaya
    • 5 years ago
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    oh i see lol

  68. myininaya
    • 5 years ago
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    your awesome

  69. anonymous
    • 5 years ago
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    you're ;p

  70. anonymous
    • 5 years ago
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    ok, I'm off to watch a movie

  71. myininaya
    • 5 years ago
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    k later peace i will see if i can come up with more challenging problems next time these sucked obviously

  72. anonymous
    • 5 years ago
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    ok =)

  73. anonymous
    • 5 years ago
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    Mod[11, 3] = 2, Mod[11, 5] = 1

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