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Classic Euler Question

  1. Oct 12, 2005 #1
    Hi
    I'm told that the the following can be presented as a classic euler question:
    A farmer goes to marked with 1770 dollars to buy rice and corn. One bag of corn costs 10 dollars and one of rice cost 20 dollar. How much corn and rice can the farmer buy?
    This can be presented in the following way?
    10x + 20y = 1770
    Then by using extended euler one can calculate x and y ???
    Regards
    Fred
     
  2. jcsd
  3. Oct 12, 2005 #2

    HallsofIvy

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    In what sense is that an "Euler question"? It looks to me like a linear Diophantine equation. Since you then refer to using the "extended euler" method, I wonder if you don't mean the "Euclidean algorithm" and "Euclidean question"? That's how I would solve this.

    Yes, you can write it as 10x+ 20y= 1770- although if I were your teacher I would insist the you first state clearly that "x is the number of bags of corn the farmer buys and y is the number of bags of rice the farmer buys".

    Of course, I assume you see immediately that 10x+ 20y= 1770 is the same as x+ 2y= 177. I might point out that 2- 1= 1 so that 1(-177)+ 2(177)= 177.

    Certainly, the farmer can't have bought -177 bags of corn! Do you notice that 2n- 177 and 177-n also satisfy 1(2n-177)+ 2(177-n)= 177?
     
    Last edited: Oct 12, 2005
  4. Jul 20, 2008 #3
    Let x be the number of bags of corn which the farmer buys.
    Let y be the number of bags of rice which the farmer buys.

    Here's a solution to the problem.
    10x + 20y = 1770 can be simplified to x+2y=177.

    Make y the subject:
    2y = 177- x .... (1)
    y = (177-x)/2 = (88*2+1-x)/2 = 44 + (1-x)/2
    Since y is an integer and 44 is an integer.
    Therefore, (1-x)/2 is an integer.

    Let (1-x) = 2a where a is any integer.
    x = 1 - 2a .... (2)

    sub. (1) into (2):
    2y = 177 - (1-2a)=176+2a
    y= 88+a

    General integer solutions are:
    x = 1-2a ..... (3)
    y = 88+a ..... (4)

    For positive integer solutions:
    Let x=1-2a > 0
    1>2a
    a<(1/2)

    Let y=88+a>0
    a>-88

    Therefore -88>a<(1/2)
    Possible values of a include: -87, -86 .... -1, 0
    There are 88 different values of a.
    ----------------------------------------------------------------
    sub. a=-1 into 3:
    x= 1-2(-1)=1+2 = 3

    sub a=-1 into 4:
    y=88+(-1) = 87

    Therefore, possible solution of the problem:
    x=3, y=87

    The farmer could have bought 3 bags of corn and 87 bags of rice.
     
  5. Jul 20, 2008 #4

    HallsofIvy

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    I started to complain that you shouldn't give complete answers to homework questions- then I noticed that this thread is 3 years old!
     
  6. Jul 20, 2008 #5
    I noticed that too.
    I saw that there hadn't been a solution to the problem and thought that it would be best to give an example of Euler's extended method, in case anyone else was interested.
     
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