## 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?

1. myininaya

or i mean what is the smaller possible value for n?

2. myininaya

smallest*

3. anonymous

Looks like the Chinese remainder theorem.

4. myininaya

yay! :)

5. myininaya

lol very good yes i used euclidean's algorithm which is mixed up with that Chinese stuff

6. myininaya

i really like this problem

7. anonymous

yeah I did something similar a few weeks ago to implement by hand RSA encryption

8. myininaya

omg polpak i love the RSA

9. myininaya

i want to be an expert at number theory

10. anonymous

pretty cool. You can work for nsa ;)

11. myininaya

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

I also looked at the secure hash algorithm 1 which is really just discrete math but still fun :)

13. myininaya

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

all algorithms are just descrete math stuff ;p

15. myininaya

yes thats true

16. myininaya

what education level are you if you don't mind me asking?

17. anonymous

About to transfer to 4 year university.

18. anonymous

just finished the local CC.

19. myininaya

what?! no way

20. myininaya

how are you super smart?

21. anonymous

I'm old. Some smarts come with age. I also study a lot (comes with being old)

22. myininaya

i think you are really cool :)

23. anonymous

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

24. myininaya

are you going to be a math person? i mean you really already are, but do something with math?

25. myininaya

math instructor? engineer? ...

26. anonymous

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

27. myininaya

you can do it you have the knowledge already for math

28. myininaya

you probably know how to solve differential equations already i think you done one here before dont really remember

29. myininaya

or some differential equations

30. anonymous

I've done 1 class in diff eq's and 1 class in linear algebra thus far

31. myininaya

are you american?

32. myininaya

americans are lazy. you can't be american because you did that self study stuff

33. anonymous

lol, yes I'm american. and yes, I'm lazy. most smart people are ;)

34. myininaya

lol

35. anonymous

I study cause it makes things easier

36. anonymous

it's a lot more work to not know how to do the homework and struggle through it for hours and hours

37. myininaya

yes i agree

38. myininaya

i do have a another challenge question for you if you are willing to accept?

39. anonymous

oh sure, though if it takes as long as euclid's algorithm I may have to hold off a while ;)

40. myininaya

$\int\limits_{}^{} e^{\sqrt{3x+9}} dx$

41. myininaya

evaluate the above is the question

42. myininaya

or the problem

43. myininaya

let me know if you want a hint k?

44. anonymous

45. myininaya

its cool

46. anonymous

looks like just u-sub then by parts..

47. anonymous

$u = \sqrt{3x + 9} \implies du = \frac{3}{u}dx \implies dx = \frac{u}{3}du$ $\implies \int e^{\sqrt{3x+9}} dx = \frac{1}{3}\int ue^udu$ Then go with by-parts.

48. anonymous

not nearly as bad as some of the triple integrals I had last year in calc 3.

49. anonymous

those things took pages =(

50. anonymous

Whoops, should be 2u/3 I think actually

51. myininaya

wait i think you can teach me something one sec

52. myininaya

53. myininaya

then integration by parts blah blah

54. anonymous

some mistakes there

55. myininaya

the only mistake i see is du=3/[2(3x+9)] dx but i really didn't use this simplification

56. anonymous

It should be $\frac{2}{3}\int ue^udu$

57. anonymous

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

oh darn it lol

59. myininaya

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

I don't square nothin

61. anonymous

I think you misunderstand what I did

62. myininaya

me too

63. myininaya

but that squaring does work also

64. anonymous

my way is easier ;p

65. anonymous

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

Then go back to the integral and sub that for dx, and the exponent of e just becomes u also.

67. myininaya

oh i see lol

68. myininaya

69. anonymous

you're ;p

70. anonymous

ok, I'm off to watch a movie

71. myininaya

k later peace i will see if i can come up with more challenging problems next time these sucked obviously

72. anonymous

ok =)

73. anonymous

Mod[11, 3] = 2, Mod[11, 5] = 1