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jagr2713

  • one year ago

Fun question :D #Bored

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  1. jagr2713
    • one year ago
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    Function F(n) defined over all positive integers n satisfies the following: \[f(1)=1,f(2)=2, and\] \[f(n+2)=f(n+1)+(f(n))^{2}+4022 \ for \ n \ge1\] How many of the 4022 integers f(1), f(2),....f(4022) are multiple of 7?

  2. myininaya
    • one year ago
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    well I notice f(3) is a multiple of 7 but checking all of those will be crazy so there has to be a slick way of doing this

  3. jagr2713
    • one year ago
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    :)

  4. dan815
    • one year ago
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    |dw:1436502484982:dw|

  5. dan815
    • one year ago
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    -1* mod 7

  6. jagr2713
    • one year ago
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    so thats the answer for you dan?

  7. dan815
    • one year ago
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    no just thinking

  8. jagr2713
    • one year ago
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    You will get it dan lol

  9. jagr2713
    • one year ago
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    hm

  10. dan815
    • one year ago
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    1

  11. dan815
    • one year ago
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    is the answer 1

  12. jagr2713
    • one year ago
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    Well nope

  13. dan815
    • one year ago
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    i think my arithmetic is wrong maybe, i got a repetition at f(9) ill put what i got

  14. jagr2713
    • one year ago
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    Lol @dan815

  15. myininaya
    • one year ago
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    well going to the whole mod thing I guess we can really just look at: \[f(n+2)=f(n+1)+(f(n))^2+4 \text{ thinking in mod } 7\] these will make the outputs easier to look at but that is still a lot of f( ) thingys to check: \[f(1)=1,f(2)=2,f(3)=7=0,f(4)=15=1, f(5)=68=5\] hmmm... \[f(6)=f(5)+(f(4))^2+4 \\ f(6)=5+1^2+4=10=3 \\ f(7)=f(6)+(f(5))^2+4 \\ f(7)=3+(5)^2+4=25=4 \\ f(8)=f(7)+(f(6))^2+4 \\ f(8)=4+(3)^2+4 \\ f(8)=1+9=10=3 \\ f(9)=f(8)+(f(7))^2+4 \\ f(9)=3+4^2+4=4^2=2 \\ f(10)=f(9)+(f(8))^2+4 \\ f(10)+2+3^2+4=6+2=8=1 \\ f(11)=f(10)+(f(9))^2+4 \\ f(11)=1+2^2+4=9=2\] \[f(12)=f(11)+(f(10))^2+4 \\ f(12)=2+1^2+4=7=0\] I think f(3) and f(12) so far are multiples of 7

  16. dan815
    • one year ago
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    |dw:1436504022956:dw|

  17. dan815
    • one year ago
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    k ya i saw myin writing same thing

  18. jagr2713
    • one year ago
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    lol nope @myininaya

  19. myininaya
    • one year ago
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    I didn't give an answer

  20. jagr2713
    • one year ago
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    Oh thought you did lol sorry

  21. myininaya
    • one year ago
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    I'm showing my work so far hoping it can help anyone else trigger more thoughts but my thought actually came from dan's work above

  22. dan815
    • one year ago
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    i have a feeling we will run into a repeating pattern in mod 7 like every 9th maybe

  23. geerky42
    • one year ago
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    I solved it by programming and I asked @jagr2713 for confirmation of my answer but he insisted me to post my answer here lol... ` numbers = [1, 2] total = 0 for x in range(3, 4023): NewNumber = (numbers[1] + (numbers[0])**2 + 4022)%7 if NewNumber%7 == 0: total +=1 numbers = [numbers[1], NewNumber] print(total) ` Output is 447

  24. ganeshie8
    • one year ago
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    Lets conjecture \[7\mid f(3+9k)\]

  25. dan815
    • one year ago
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    :P 4022 / 9 = 446.8 pretty close hehe

  26. ganeshie8
    • one year ago
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    based on @myininaya 'w work..

  27. jagr2713
    • one year ago
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    WOw great work

  28. jagr2713
    • one year ago
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    Great job :D

  29. dan815
    • one year ago
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    brute force cheattoorr

  30. jagr2713
    • one year ago
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    lol dan got right after geerky lol

  31. myininaya
    • one year ago
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    I see pattern 1,2,0,1,5,3,4,3,2,1,2,0,1,5,3,4,3,2,1,2,0,1,5,3,4,3,2,1,2,0,1,5,3,4,3,2,... I wonder if this is the pattern we have a 0 at 3rd term,12th term, 21th term, oh and I see where @ganeshie8 get's his conjecture now

  32. dan815
    • one year ago
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    yep

  33. dan815
    • one year ago
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    ask another one now!

  34. jagr2713
    • one year ago
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    k let start typing it :D

  35. ganeshie8
    • one year ago
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    the proof might be interesting

  36. dan815
    • one year ago
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    by induction it can be proven

  37. myininaya
    • one year ago
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    I would love to see a proof on this but I think that is above me.

  38. dan815
    • one year ago
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    once you get a full repettiion it must be true

  39. dan815
    • one year ago
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    but there could be a neater proof this is a lame way to prove it

  40. ganeshie8
    • one year ago
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    that is true, once the remainders start repeating, we're technically done!

  41. myininaya
    • one year ago
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    sadly i can't stay for the fun @jagr2713 will post it is time for me to retire :(

  42. dan815
    • one year ago
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    good night myin

  43. jagr2713
    • one year ago
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    For all real numbers x, the function f(x) satisfies \[f(1+x)+xf(1-x)=x ^{2}+9x\] If the values of f(-7) can be expressed as \[\frac{ a }{ b }\] Where a and b are coprime positive integers/ What is a+b Here you go @dan815

  44. jagr2713
    • one year ago
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    see ya @myininaya D:

  45. dan815
    • one year ago
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    f(1+x)=9x f(1-x)=x?

  46. dan815
    • one year ago
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    x*f(1-x)= x^2+ax+b f(1+x)=cx+d

  47. dan815
    • one year ago
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    oh and since u applied x after there can be no constant term in both

  48. dan815
    • one year ago
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    therefore f(1-x)=x+b and f(1+x)=ax they must of this tform

  49. dan815
    • one year ago
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    f(1+x) cannot have a constant term

  50. dan815
    • one year ago
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    and f(1-x) cannot have a coefficient on the x

  51. dan815
    • one year ago
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    so we are restrited with the unknowns here

  52. dan815
    • one year ago
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    restricted*

  53. jagr2713
    • one year ago
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    hm

  54. dan815
    • one year ago
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    f(1-x)=x+b and f(1+x)=ax f(-7)=-7+b f(-7)=-7a and f(-7)= c/d, where c and d are co prime int hmm i am wondering if the first lines are really okay to assume

  55. jagr2713
    • one year ago
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    hm lol waiting lol falling asleep lol

  56. dan815
    • one year ago
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    f(-7)=-7+b f(-7)=-7a a+b=9 f(-7)=-7+ 9-a =2-a f(-7)=-7*(a) 2-a=-7a a=1/3

  57. dan815
    • one year ago
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    lol what is going on

  58. Empty
    • one year ago
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    I don't know but I'll guess 13 and then I think I'll try solving it to make sure.

  59. ganeshie8
    • one year ago
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    \[\large a+b=217\]

  60. dan815
    • one year ago
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    howd u do it

  61. jagr2713
    • one year ago
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    Nope ganesh D: is that your answer @dan815

  62. jagr2713
    • one year ago
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    but your close

  63. ganeshie8
    • one year ago
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    simply plugin \(x=\pm 8\), you will get two linear eqn's which you can solve

  64. ganeshie8
    • one year ago
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    \[\large a+b=229\] ?

  65. jagr2713
    • one year ago
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    Correcto :D Finally now i can sleep lol

  66. jagr2713
    • one year ago
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    dont know who to give the stars D:

  67. Empty
    • one year ago
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    Ahhh good trick @ganeshie8 ! I had tried to turn this into a differential equation like yesterday and got something completely terrible.

  68. Empty
    • one year ago
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    Ahhh good trick @ganeshie8 ! I had tried to turn this into a differential equation like yesterday and got something completely terrible.

  69. dan815
    • one year ago
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    ah i liked it

  70. Empty
    • one year ago
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    It looks like you also get the answer to what f(9) is for free as well haha

  71. ganeshie8
    • one year ago
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    I subed 1+x=u and 1-x=v and tried to get a differential eqn like yesterday but got stuck

  72. Empty
    • one year ago
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    @ganeshie8 What you could try instead is plug in 1-x=y+h and 1-x=y and that'll turn it into a differential equation.

  73. ganeshie8
    • one year ago
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    this could be an interesting question For all real numbers x, the function f(x) satisfies \[f(1+x)+xf(1-x)=x ^{2}+9x\] find \(f(x)\) Here you go @dan815

  74. ganeshie8
    • one year ago
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    Ahh okay I see that should make it easy..

  75. Empty
    • one year ago
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    Unfortunately (I might have done it wrong) the differential equation we get is not a very pretty one to solve but we can do it with an integrating factor it seems, I just think it ends up having the erf(x) function in there as part of the answer #_#

  76. dan815
    • one year ago
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    do you think this assumption is right f(1+x)=ux f(1-x)=x+v

  77. Empty
    • one year ago
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    Hmmm I don't know?

  78. dan815
    • one year ago
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    if we ignore the fourier series and other orthogonal representations to polynomial functions

  79. dan815
    • one year ago
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    f(1+x)+x*f(1-x) = x^2+-9x then can we say f(1+x) must be 1 degree lower than x*f(1-x) so the x^2 term must come from that term

  80. Empty
    • one year ago
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    Ahhh I see what you're looking at now and why you would say that but I'm sure if I know if we can do that or not. Interesting observation though. It looks like we should be able to solve for f(x)=ax+b by plugging in I think.

  81. ganeshie8
    • one year ago
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    yeah at least we will get all linear functions, if not all

  82. dan815
    • one year ago
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    yea because if this is really true u get restrictions all over the place since then x*f(1-x) cannot have a constant term that means f(1-x) must have form ax+b and then f(1+x) cannot have a constnat term as it all equalys x^2+9x

  83. dan815
    • one year ago
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    f(1-x)= x + u f(1+x)=v*x u+v=9

  84. Empty
    • one year ago
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    I was also considering if we get anything by splitting it up into an even and odd function: f(x)=g(x)+h(x) g(x)=g(-x) and h(x)=-h(-x) since we can always do this, then make a pair of equations this way that will be easy to solve hopefully.

  85. dan815
    • one year ago
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    oo

  86. ganeshie8
    • one year ago
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    Looks f(x) cannot be a polynomial

  87. ganeshie8
    • one year ago
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    does this work f(1+x) + xf(1-x) = x^2+9x f(1-x) -xf(1+x) = x^2-9x ----------------------------- f(1-x)(1+x^2) = x^3+10x^2-9x f(1-x) = (x^3+10x^2-9x)/(1+x^2)

  88. dan815
    • one year ago
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    i think so, now u can rewrite the right side as functions of 1-x ?

  89. ganeshie8
    • one year ago
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    wow yeah 1-x = u should do

  90. Empty
    • one year ago
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    Nice, that wasn't so bad after all haha

  91. Empty
    • one year ago
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    So if we look at f(x) as being a generating function, what's the sequence it generates? :O

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