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  1. #1
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    word problem

    "The sum of the first two digits of a three-digit number exceeds the third digit by 8. If the first two digits are interchanged the new number exceeds the original number by 270. If the number is divided by the sum of the digits the quotient is 31 and the remainder 15. Find the original number."
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  2. #2
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    Quote Originally Posted by biva View Post
    "The sum of the first two digits of a three-digit number exceeds the third digit by 8. If the first two digits are interchanged the new number exceeds the original number by 270. If the number is divided by the sum of the digits the quotient is 31 and the remainder 15. Find the original number."
    Let the number be "xyz." That is to say the number is
    100x + 10y + z

    We know that
    x + y = z + 8

    100y + 10x + z = (100x + 10y + z) + 270

    \frac{100x + 10y + z}{x + y + z} = 31 + \frac{15}{x + y + z}

    Start with the top and work your way down:
    z = x + y - 8

    Inserting this into the other two expressions gives:
    100y + 10x + (x + y - 8) = (100x + 10y + (x + y - 8)) + 270 ==> 101y + 11x - 8 = 101x + 11y + 262

    and
    \frac{100x + 10y + (x + y - 8)}{x + y + (x + y - 8)} = 31 + \frac{15}{x + y + (x + y - 8)} ==> \frac{101x + 11y - 8}{2x + 2y - 8} = 31 + \frac{15}{2x + 2y - 8}

    Let's look at the new top equation:
    101y + 11x - 8 = 101x + 11y + 262

    -90x + 90y = 270

    90y = 90x + 270

    y = x + 3

    Inserting this into the new bottom equation:
    \frac{101x + 11(x + 3) - 8}{2x + 2(x + 3) - 8} = 31 + \frac{15}{2x + 2(x + 3) - 8}

    \frac{112x + 25}{4x - 2} = 31 + \frac{15}{4x - 2}<-- Multiply both sides by 4x - 2

    112x + 25 = 31(4x - 2) + 15

    112x + 25 = 124x - 62 + 15

    -12x + 25 = - 47

    -12x = -72

    x = 6

    So
    y = x + 3 ==> y = 9

    and

    z = x + y - 8 ==> z = 7

    So your number is 697.

    -Dan
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  3. #3
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    Hello, biva!

    Another approach . . . with fewer fractions.


    The sum of the first two digits of a three-digit number exceeds the third digit by 8.
    If the first two digits are interchanged, the new number exceeds the original number by 270.
    If the number is divided by the sum of the digits, the quotient is 31 and the remainder 15.
    Find the original number.
    Let the digits be x,\,y,\,z.
    The three-digit number is: 100x + 10y + z

    The first sentence says: . x + y \:=\:z + 8\quad\Rightarrow\quad x + y - z \:=\:8 .[1]

    The second says: . 100y + 10x + z \:=\:100x+ 10y + z + 270\quad\Rightarrow\quad x - y \:=\:-3 .[2]

    The third says: . \frac{100x + 10y + z}{x + y + z} \:=\:31 + \frac{15}{x + y + z}\quad\Rightarrow\quad23x - 7y - 10z \:=\:5 .[3]

    We have a system of equations.


    From [2], we have: . y \:=\:x + 3 .[4]

    Substitute into [1]: . x + (x + 3) - z \:=\:8\quad\Rightarrow\quad 2x - z\:=\:5 .[5]

    Substitute into [3]: . 23x - 7(x+3) - 10z \:=\:5\quad\Rightarrow\quad 8x - 5z \:=\:13 .[6]

    Multiply [5] by -5: . \text{-}10x + 5x \:=\:\text{-}25
    . . . . . . .Add [6]: . . 8x - 5x \:=\;13

    . . and we get: . -2x \:=\:-12\quad\Rightarrow\quad\boxed{x \,=\,6}

    Substitute into [4]: . y \:=\:6 + 3\quad\Rightarrow\quad\boxed{y \,=\,9}

    Substitute into [5]: . 2(6) - z \:=\:5\quad\Rightarrow\quad\boxed{z \,=\,7}


    Therefore, the original number is
    697
    .
    Last edited by Soroban; June 22nd 2007 at 11:45 AM.
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