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Math Help - Difficult Counting Problem

  1. #1
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    Difficult Counting Problem

    What is the number of positive integers n satisfying the property that the number b of positive integers a satisfying the property that n a and
    a^2 n satisfies the property that the number of possible ways we can put 3 different objects into b different boxes is at least as big as the number of
    0-1 sequences of length n.

    Thanks for your help!
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  2. #2
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    So I have attempted a solution... Does this make sense? The question is incredibly confusing wording but here are my thoughts to the problem...

    RELEVANT EQUATIONS:
    3 different objects into b different boxes = b^3
    number of 0,1 sequences of length n = 2^n

    SOLUTION ATTEMPT:
    From the question, we know n is between a and a^2.
    n is the number of positive integers
    a is the positive integers before n
    b is the number of a
    and we know that b^3 is greater or equal to 2^n
    if we try n=1, then a=1, b=1
    n=2, then a=1,2, b=2
    n=3, then a=1,2,3, b=3...
    therefore n=b, which gives n^3 is greater than or equal to 2^n.
    I don't know how to get n in other way, so I tried until n=10.
    if n=10, then a=1,2,3,4,5,6,7,8,9,10, b=10
    then 10^3 is not greater or equal to 2^10.
    Therefore n is between 2 and 9 since if n=1, then a=1, b=1
    1^3 is not greater of equal to 2^1.
    Answer is 8.
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  3. #3
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    If my understanding is correct, then n must be less than a^2. However your cases where you list the possible a's don't include this case. Therefore, although it is true for a=1, b is not equal to n.

    I have done trial and error and have found the answer to be 3.

    Cases:
    N=1 a <= 1 <= a^2
    a = 1, b =1

    N=2 a <= 2 <= a^2
    A=2, b = 1
    N=3 a <= 3 <=a^2
    A = 2, 3, b=2
    N=4 a <= 4 <= a^2
    A = 2,3,4 b=3
    N = 5 a <= 5 <= a^2
    A= 3,4,5 b = 3
    N = 6 a <= 6 <= a^2
    A = 3,4,5,6 b = 4
    N = 7 a<= 7 <= a^2
    A = 3,4,5,6,7 b=5
    N = 8 a<= 8 <= a^2
    A = 3,4,5,6,7,8 b = 6
    N = 9 a<= 9 <= a^2
    A = 3,4,5,6,7,8,9 b= 7
    N = 10 a<=10 <= a^2
    A = 4,5,6,7,8,9,10 b=7
    etc...

    I think this is the solution if n is between a and a^2. Does this make sense to everyone? It's late here and I may have no idea what I'm doing. What does everyone else think?
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