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Sum of power of two exact 2010 different ways

  1. Nov 27, 2009 #1
    Hello,

    I look for a solution for the following problen.

    Determine all numbers, which can be written in exact 2010 different ways as a sum of powers of two with non-negative exponent, while all exponents are only allowed to appear maximal three times in one sum.
    (All sums where only the order of the summands is changed are count as one sum.)


    My problem is that I do not have any idea, how to find a solution to this task.
    Therefore I started to determine in how many different sums I can write for example 2^4:
    1. 2^4
    2. 2*2^3
    3. 2*2^2+2^3
    4. 2*2+2^2+2^3
    5. 2*2^0+2+2^2+2^3
    6. 2*2^0+3*2+2^3
    7. 2*2^0+3*2+2*2^3

    Nevertheless, I did not find any coherenz zu the number and the number of different sums.

    Do you have any tips?

    Christian
     
  2. jcsd
  3. Nov 30, 2009 #2
  4. Nov 30, 2009 #3

    Mark44

    Staff: Mentor

    I don't understand what you're asking. Is it that you want all numbers that can be written in exactly 2010 different ways?

    Or is it that you want to find the number of ways that 2010 can be written as a sum of powers of 2?
     
  5. Dec 1, 2009 #4
    Hi there!
    I believe you quote the question from Columbus state university .

    This is an easy question.
    Assume there is four blocks since the two non zero integer are similar ,
    we label one of the a and b. one with a selection of 3 integers another with 4.
    For a>2.


    Therefore the blocks of 4 integer will look like this:
    a00b
    a0b0
    ab00

    I'll leave the rest to you.
    ans is thirty-_____ .
     
  6. Dec 1, 2009 #5
    Huh? You're sure this is the answer to the right question?
     
  7. Dec 1, 2009 #6
    there are actually 9 ways to write 2^4

    writing them in 'binary'

    10000
    2000
    1200
    1120
    1112
    1032
    320 = 3*2^2 + 2 * 2^2
    312 = 3*2^2 + 1 * 2^1 + 2*2^0
    232 = 2*2^2 + 3 * 2^1 + 2*2^0

    you missed the last 3 and your number 7 isn't valid.

    It appears that if f(n) is the number of ways to write n, then f(n) = 1 + floor(n/2)

    (floor(x) is the greatest integer <= x)

    I haven't been able to prove this so far. It's easy to prove that f(2n+1) = f(2n) so you
    only have to prove f(n) = 1+n/2 for even numbers.
     
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