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Math Help - Extremly hard question! Its a Challenge!

  1. #1
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    Extremly hard question! Its a Challenge!

    aa
    Last edited by rafilv; April 9th 2008 at 02:19 PM.
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  2. #2
    Super Member angel.white's Avatar
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    let the digits be represented by letters. Then the palindrome is
    abcdefghhgfedcba

    For multiplication this means a^2 b^2 c^2 d^2 e^2 f^2 g^2 h^2 = (a b c d e f g h)^2 = 16

    then abcdefgh = 4

    possible solutions:
    one letter = 4, up to 7 of the others equal 1
    two letters equal 2, up to 6 of the others equal 1



    For addition this means 2a+2b+2c+2d+2e+2f+2g+2h = 2(a+b+c+d+e+f+g+h) = 16

    then a+b+c+d+e+f+g+h = 8

    We know that they both equal 16 so possible solutions for addition from the set of possible solutions above (the intersection of their solution sets):
    one letter equals 4, 4 letters equal 1, 3 letters equal zero
    2 letters equal 2, 4 letters equal 1, 2 letters equal zero



    Now we just need to find how many permutations there are in this set for the two solutions, then add them together. I'm not entirely sure how to do that.

    My best attempt would look like this:
    There are 8 possibilities for a 4, 7 for a 1, 6 for a 1, 5 for a 1, 4 for a 1, the rest are zero so 8*7*6*5*4 = 6720

    There are 8 possibilities for a 2, 7 possibilities for a 2, 6 possibilities for a 1, 5 possibilities for a 1, 4 possibilities for a 1, 3 possibilities for a 1, the rest are zero so 8*7*6*5*4*3 = 20160

    Their sum is 6720 + 20160 = 26880 different ways of making the palindrome.



    I think it is correct up until the calculating of probability, but after that point I don't have much confidence.
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  3. #3
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    Quote Originally Posted by angel.white View Post
    We know that they both equal 16 so possible solutions for addition from the set of possible solutions above (the intersection of their solution sets)ne letter equals 4, 4 letters equal 1, 3 letters equal zero
    2 letters equal 2, 4 letters equal 1, 2 letters equal zero
    I think that the above answer is a gross over count. Although part of the analysis is right on.

    Any 16-dight palindrome is completely determined by the first eight digits.
    Because 16=(4)(4)=(8+8), it is clear that the sum must be 8 and the product of the non-zero digits must be 4.
    Therefore as was found above, there are two possiblies: 41111000 or 22111100.
    So the answer is the sum of the number of ways to rearrange each of those strings that do not begin with a zero.
    Rearrangements of the string "41111000" can begin with 4 in \frac{{7!}}{{\left( {4!} \right)\left( {3!} \right)}} ways and can begin with a 1 in \frac{{7!}}{{\left( {3!} \right)^2 }}.

    Rearrangements of the string "22111100" can begin with a 2 in \frac{{7!}}{{\left( {4!} \right)\left( {2!} \right)}} ways and can begin with a 1 in \frac{{7!}}{{\left( {3!} \right)\left( {2!} \right)^2 }}.

    What is that sum?
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  4. #4
    Super Member angel.white's Avatar
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    Quote Originally Posted by Plato View Post
    I think that the above answer is a gross over count. Although part of the analysis is right on.

    Any 16-dight palindrome is completely determined by the first eight digits.
    Because 16=(4)(4)=(8+8), it is clear that the sum must be 8 and the product of the non-zero digits must be 4.
    Therefore as was found above, there are two possiblies: 41111000 or 22111100.
    So the answer is the sum of the number of ways to rearrange each of those strings that do not begin with a zero.
    Rearrangements of the string "41111000" can begin with 4 in \frac{{7!}}{{\left( {4!} \right)\left( {3!} \right)}} ways and can begin with a 1 in \frac{{7!}}{{\left( {3!} \right)^2 }}.

    Rearrangements of the string "22111100" can begin with a 2 in \frac{{7!}}{{\left( {4!} \right)\left( {2!} \right)}} ways and can begin with a 1 in \frac{{7!}}{{\left( {3!} \right)\left( {2!} \right)^2 }}.

    What is that sum?
    35 + 140 + 105 + 210 = 490
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