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Barcallica

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Posts posted by Barcallica

  1. Host: Molly Mae

    1. Flamebirde - 2013
    2. MikeD - 2013
    3. DyalDragon
    4. Kikacat123 - 2013
    5. Marksmanjay
    6.
    7. Barcallica - 2013
    8. Brainiac100
    9.
    10. araver - 2013
    11.
    12. dee_tot - 2013
    13. TwoaDay
    14.
    15.Magic - 2013
    16. Hirkala
    17. Panther - 2013
    18. Bmad - 2013
    19. mboon

    And now for something completely different... Hope I can join in.

  2. what we are doing on the right side is what we do to convert from decimal to binary. So what we are trying to prove is "if n in decimal equals a

    1a2....ak in binary then prove in decimal n=a1*2k-1+a2*2k-2......+ak-1*2+ak, which is another way we convert dec to bin. Maybe someone help me proving this conversion methods???

    this is one approach

    i found it easier to prove it geometrically

    Interesting, I have no idea how you could connect it to geometry.

  3. what we are doing on the right side is what we do to convert from decimal to binary. So what we are trying to prove is "if n in decimal equals a

    1a2....ak in binary then prove in decimal n=a1*2k-1+a2*2k-2......+ak-1*2+ak, which is another way we convert dec to bin. Maybe someone help me proving this conversion methods???
  4. Effectively, you are adding the numbers on the left. Now, all the numbers on the left are doubles (even) except the first number that you start with. So, irrespective of the number you have on the right, the addition on the left will be odd if you start with an odd number on the left.

    Therefore, this algorithm is not correct.

    That is not true.

    if the first number number on the right is even then that row will be crossed, leaving sum of left column even not odd. So it works in every possible pair.

    Actually this has nothing to do with the number on the left. And also has something to with the fact that every number can be put as sum of exponents of 2.

  5. Simplifying all equations without any solutions to make things easier:

    nn-1/2

    sqrt(n)-1

    sqrt(n)+1

    n2

    (3n)!-n

    (3(n!))!+n

    2srqt(n)+1

    n*1n (=n if n<infinity)

    2+n

    2n+1

    2n+(n-1)!

    2n+|n| (=3n, n>0, or =n, n<0)

    2n*.n+.n (=.n*(2n+1))

    2n

    (n+1)/.n

    2n-(sqrt(n)-1)!

    2n+1

    2nn+n

    (n-1)!-n

    2n!-1

    n*sqrt((n-.n)/.n)

    n(1+2/.n)

    sqrt(.n-n)*n-sqrt(n)

    (nn/n-n)!

    n2

    I thought it was more of a fill-in thing, that squares

  6. For every round you have chance to get average 1/2$. More detailed, you have chance to lose 5/8, chance to recover 15/56, chance to win 1$ 5/56, to win 2$ is 1/56

    Bonus: You guys will collectively lose 3$ every round to the house

    How much does a player stand to lose per round they play?

    I didnt quite understand the question. But Curtis Jackson??

  7. For every round you have chance to get average 1/2$. More detailed, you have chance to lose 5/8, chance to recover 15/56, chance to win 1$ 5/56, to win 2$ is 1/56



    Bonus: You guys will collectively lose 3$ every round to the house

  8. Let a1, a2, ....am be the sides of the cakes. we need to prove 4*(a1+a2+......+am)<=20*sqrt(mn) ------> (a1+a2+......+am)<=5*sqrt(mn) -------> (a1+a2+......+am)2<=25mn We know 25n>=a12+a22+.........+am2 so we need to prove m*(a12+a22+.........+am2)>=(a1+a2+......+am)2 which can be easily proven by writing equation a2+b2>=2ab; (m-1)*(m-2)/2 times.

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