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Math Help - Rod pRoblems

  1. #1
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    Exclamation Rod pRoblems

    A carpenter has a long rod of length 4m 38 cm. He needs to cut this rod into lengths of 161cm, 23cm and 7cm so that he would get at least one rod of each length and have nothing left over. After a while, the carpenter managed to solve the problem. How many rods of each length did he get?
    Find all possible solutions.

    Whoever does this is my heroe
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  2. #2
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    Quote Originally Posted by googa View Post
    A carpenter has a long rod of length 4m 38 cm. He needs to cut this rod into lengths of 161cm, 23cm and 7cm so that he would get at least one rod of each length and have nothing left over. After a while, the carpenter managed to solve the problem. How many rods of each length did he get?
    Find all possible solutions.

    Whoever does this is my heroe
    Here is one way.

    4m 38cm = 438cm

    How many 161 cm are in 438 cm?
    Two
    So, what is left is 438 -2(161) = 116 cm

    Let x = number of 23-cm rods
    And y = number of 7-cm rods
    So,
    23x +7y = 116
    x = (116 -7y)/23 -----**

    By repetition, find the the integer value of y that will produce an integer x.
    I found y = 10 to give x = 2
    That means 10 rods of 7cm long
    And 2 rods of 23 cm long

    Therefore, the carpenter got 2 rods of 161cm long, 2 rods of 23cm long, and 10 rods of 7cm long. -------------answer.

    Check:
    2(161) +2(23) +10(7) =? 438
    438 =? 438
    Yes, so, OK.
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  3. #3
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    let x = number of 161 lengths
    let y = number of 23 lengths
    let z = number of 7 lengths



    Then 161x + 23y + 7z = 438

    Note x can only be 1 or 2

    if x = 1 then 23y + 7z = 277

    substitute all possible values of y (ie 11,10,...,1) . select values which give whole number solution for z.


    similarly if x = 2 then 23y + 7z = 116

    again substitute all possible y values (ie 4,3,2,1 ). select values which give a whole number solution for z
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  4. #4
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    Hello, googa!

    Yet another approach . . .


    A carpenter has a long rod of length 438 cm.
    He needs to cut this rod into lengths of 161cm, 23cm and 7cm
    so that he would get at least one rod of each length and have nothing left over.
    After a while, the carpenter managed to solve the problem.

    How many rods of each length did he get?
    Find all possible solutions.
    I found three solutions.


    Let: . \begin{array}{ccc}L &=&\text{number of 161-cm rods} \\ M &=& \text{number of 23-cm rods} \\ S &=& \text{number of 7-cm roads} \end{array}


    We have: . 161L + 23M + 7S \:=\:438 \quad\Rightarrow\quad S \;=\;\frac{438 - 161L - 23M}{7}

    . . Hence: . S \;=\;62 - 23L - 4M + \frac{4+5M}{7}


    Since S is an integer, 4 + 5M must be divisible by 7.
    . . This happens for M \:=\:2,9,16,\hdots


    If M = 2, we have: . S \:=\:56-23L . . . then: . \begin{array}{cc}L \:=\:1, & S \:=\:33 \\ L \:=\:2, & S \:=\:10 \end{array}

    . Two solutions: . (L,M,S) \:=\:\boxed{(1,2,33),\;(2,2,10)}


    If M = 9, we have: . S \:=\:33-23L . . . then: . L = 1,\;\;S = 10

    . . One solution: . (L,M,S) \:=\:\boxed{(1,9,10)}


    If M = 16, we have: . S \:=\:10-23L . . . no solutions

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