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Math Help - [SOLVED] Difficult binomial theorem question(involving greatest integer function)?

  1. #1
    Super Member fardeen_gen's Avatar
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    [SOLVED] Difficult binomial theorem question(involving greatest integer function)?

    If (3√3 + 5)^9 = R and [R] denotes the greatest integer less than or equal to R, then:

    A) [R] is divisible by 10
    B) [R^2] is divisible by 512
    C) [R] is divisible by 15
    D) [R] is an even number


    More than one options may be correct.
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  2. #2
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    Binomial

    Hello again fardeen_gen

    As with your previous posting, I'm not sure how you're supposed to tackle this question. I have used a calculator, and found [R] to be 1191041440. So A and D are true. The others are not.

    Grandad
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    Quote Originally Posted by fardeen_gen View Post
    If (3√3 + 5)^9 = R and [R] denotes the greatest integer less than or equal to R, then:

    A) [R] is divisible by 10
    B) [R^2] is divisible by 512
    C) [R] is divisible by 15
    D) [R] is an even number


    More than one options may be correct.
    Here is another approach:

    (3 \sqrt{3} + 5)^9 = \sum_{i=0}^9 \binom{9}{i} 3^{3(9-i)/2} 5^i
    (3 \sqrt{3} - 5)^9 = \sum_{i=0}^9 (-1)^i \binom{9}{i} 3^{3(9-i)/2} 5^i

    Subtracting, the even-numbered terms cancel, yielding

    (3 \sqrt{3} + 5)^9 - (3 \sqrt{3} - 5)^9 = 2 \sum_{i=0}^4 \binom{9}{2i+1} 3^{3(4-i)} 5^{2i+1}

    So
    (3 \sqrt{3} + 5)^9  = (3 \sqrt{3} - 5)^9 + 2 \sum_{i=0}^4 \binom{9}{2i+1} 3^{3(4-i)} 5^{2i+1}

    [(3 \sqrt{3} + 5)^9] = 2 \sum_{i=0}^4 \binom{9}{2i+1} 3^{3(4-i)} 5^{2i+1}
    because
    0 < 3 \sqrt{3} - 5 < 1 and the summation is an integer.
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  4. #4
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    Binomial

    Quote Originally Posted by awkward View Post
    Here is another approach:

    (3 \sqrt{3} + 5)^9 = \sum_{i=0}^9 \binom{9}{i} 3^{3(9-i)/2} 5^i
    (3 \sqrt{3} - 5)^9 = \sum_{i=0}^9 (-1)^i \binom{9}{i} 3^{3(9-i)/2} 5^i

    Subtracting, the even-numbered terms cancel, yielding

    (3 \sqrt{3} + 5)^9 - (3 \sqrt{3} - 5)^9 = 2 \sum_{i=0}^4 \binom{9}{2i+1} 3^{3(4-i)} 5^{2i+1}

    So
    (3 \sqrt{3} + 5)^9  = (3 \sqrt{3} - 5)^9 + 2 \sum_{i=0}^4 \binom{9}{2i+1} 3^{3(4-i)} 5^{2i+1}

    [(3 \sqrt{3} + 5)^9] = 2 \sum_{i=0}^4 \binom{9}{2i+1} 3^{3(4-i)} 5^{2i+1}
    because
    0 < 3 \sqrt{3} - 5 < 1 and the summation is an integer.
    That is pretty neat! It's clear, then, that R is a multiple of 10. And it's not a multiple of 3, since all the terms are except the last. Therefore R is not a multiple of 15. Any idea how this might determine the truth of statement B? (Other than by evaluating [R], of course.)

    Grandad

    PS I have just looked again at statement B, and noticed that it is R^2, not just R. Since 512 = 2^9, this will be true if R is a multiple of 32. And it is. So B is true after all!
    Last edited by Grandad; February 14th 2009 at 10:08 PM.
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