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Math Help - Problem Of The Week

  1. #1
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    Problem Of The Week

    Find an integer between 100 and 200 such that each digit is odd and the sum of the cubes of the digit is equal to the original three-digit number. Hint: the "cube" of a number is that number multiplied by itself 3 times, so the cube of 2 is 2*2*2=8
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    Hello, aznmartinjai!

    Find an integer between 100 and 200 such that each digit is odd
    and the sum of the cubes of the digits is equal to the original three-digit number.
    With a little thinking, you can narrow down the choices.

    Since the number is between 100 and 200, the first digit is "1".

    The other digits must be odd: 1, 3, 5, 7, 9

    But 7 = 343 and 9 = 729 . . . too large.

    So the number is comprised of 1s, 3s, and/or 5s.


    If the number has no 5s, the largest is 133 . . 1 + 3 + 3 .= .55
    . . Hence, the number contains one 5.


    There are only four choices: 115, 135, 151, 153.

    . . 115 . . 1 + 1 + 5 .= .127

    . . 135 . . 1 + 3 + 5 .= .153

    . . 151 . . 1 + 5 + 1 .= .127

    . . 153 . . 1 + 5 + 3 .= .153 . .There!

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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Soroban View Post
    Hello, aznmartinjai!

    With a little thinking, you can narrow down the choices.

    Since the number is between 100 and 200, the first digit is "1".

    The other digits must be odd: 1, 3, 5, 7, 9

    But 7 = 343 and 9 = 729 . . . too large.

    So the number is comprised of 1s, 3s, and/or 5s.


    If the number has no 5s, the largest is 133 . . 1 + 3 + 3 .= .55
    . . Hence, the number contains one 5.


    There are only four choices: 115, 135, 151, 153.

    . . 115 . . 1 + 1 + 5 .= .127

    . . 135 . . 1 + 3 + 5 .= .153

    . . 151 . . 1 + 5 + 1 .= .127

    . . 153 . . 1 + 5 + 3 .= .153 . .There!

    don't mean to be a pain, but your method works out nicely since we were dealing with a relatively short range. how would you construct a "method" to get the answer if the range was too large for trial and error?
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    Quote Originally Posted by Jhevon View Post
    don't mean to be a pain, but your method works out nicely since we were dealing with a relatively short range. how would you construct a "method" to get the answer if the range was too large for trial and error?
    It is going to lead to the following Diophantine Equation:

    x^3+y^3+z^3 = 100x + 10y +z

    I think it is safe to safe, there is no nice algorithm to this equation.
    This is very typical with diophantine equations.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    It is going to lead to the following Diophantine Equation:

    x^3+y^3+z^3 = 100x + 10y +z

    I think it is safe to safe, there is no nice algorithm to this equation.
    This is very typical with diophantine equations.
    i see that! so in this particular case, x will always be 1 and y,z will be either 3 or 5, so Soroban just found the specific case
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