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Math Help - Diophantine equation

  1. #1
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    Question Diophantine equation

    Question:

    Find all solutions that are positive integers of the Diophantine equation:

    25x + 35y = 380

    I need to check my answer. Not sure If I have the correct solutions

    x > 0
    y > 0

    25x + 35 = 380 ==> x = 13.8

    25 + 35y = 380 ==> y = 10.14

    0 < x < 14
    0 < y < 11

    25x + 35y = 380 ==> 5x + 7y = 76
    25x + 175k + 105 = 380 making
    x = 11-7k


    25(11) + 35(4) = 275 + 105 = 380
    25(4) + 35(8) = 100 + 280 = 380

    Solution
    K(0,1)
    X(11,4)
    Y (3,8)


    Not sure if I expressed my answer corretly.
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  2. #2
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    Quote Originally Posted by math_cali View Post
    Question:

    Find all solutions that are positive integers of the Diophantine equation:

    25x + 35y = 380
    Now, this is actually the way you can get the answer to the other problem you posted without guessing.
    ---
    There is a cool way to do these using continued fractions but I do not think you number theory course does those until the end of the semester if not at all.

    We need to find,
    \gcd(25,35)
    Apply Euclidean algorithm,
    35=1\cdot 25+10
    25=2\cdot 10+5
    10=2\cdot 5 + 0
    Now work backwards,
    5=25-2\cdot 10
    5=25-2(35-25)
    5=25-2\cdot 35+2\cdot 25
    5=3\cdot 25-2\cdot 35
    Thus,
    25(3)+35(-2)=5
    Multiply by 76,
    25(228)+35(-152)=380
    Thus,
    x=228+(35/5)t=228+7t
    y=-152-(25/5)t=-152-5t
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  3. #3
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    oh ok thanks alot I thought I was suppost to be looking for exact numbers. Now I understand and see how I could apply this in the previous problem I posted
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  4. #4
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    Hello, math_cali!

    Find all solutions that are positive integers of the Diophantine equation:
    . . 25x + 35y \:= \:380

    Your answers are correct!
    I would write them as: . (x,y) \:=\:(4,8),\:(11,3)

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  5. #5
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    I kind of understand how I got the solutions when I wrote it out the frist time, but how would you apply the solutions to what ThePerfectHacker showed me when he worked it out?
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  6. #6
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    Hello, math_cali!

    Here's a really primitive approach . . .


    Find all positive integer solutions of: . 25x + 35y \:= \:380

    Divide by 5: . 5x + 7y \:=\:76

    Solve for x\!:\;\;x \:=\:\frac{76-5y}{5}

    This can be written: . x \:=\:\frac{75+1-10y+3y}{3} \:=\:\frac{75 - 10y + 1 + 3y}{5}\:=\:\frac{75-10y}{5} + \frac{1 + 3y}{5}

    . . Hence: . x \:=\:15 - 2y + \frac{1+3y}{5} [1]


    Since x is an integer, \frac{1+3y}{5} must some integer a.
    . . We have: . \frac{1+3y}{5}\:=\:a\quad\Rightarrow\quad y \:=\:\frac{5a-1}{3}

    This can be written: . y \:=\:\frac{3a+2a-3+2}{3}\:=\:\frac{3a-3+2a+2}{3}\:=\:\frac{3a-3}{3} + \frac{2a+2}{3}

    . . Hence: . y \:=\:a - 1 + \frac{2a+2}{3} [2]


    Since y is an integer, then \frac{2a+2}{3} must be some integer b.

    . . We have: . \frac{2a+2}{3}\:=\:b\quad\Rightarrow\quad a\:=\:\frac{3b-2}{2}

    . . Hence: . a\:=\:\frac{3b}{2} - 1 [3]


    Since a is an integer, b must be even: . b\,=\,2k

    Substitute into [3]: . a \:=\:\frac{3(2k)}{2} - 1\quad\Rightarrow\quad a \:=\:3k-1

    Substitute into [2]: . y \:=\:(3k-1) - 1 + \frac{2(3k-1) + 2}{3}\quad\Rightarrow\quad y\,=\,5k-2

    Substitute into [1]: . x\:=\:15 - 2(5k-2) + \frac{1 + 3(5k-2)}{5}\quad\Rightarrow\quad x \:=\:18-7k


    We have found all solutions: . \begin{Bmatrix}x\:=\:18 - 7k \\ y \:=\:5k-2\end{Bmatrix} for any integer k.


    Since x and y must be positive integers:
    . . \begin{array}{cc}18 - 7k\:>\:0 \\ 5k-2\:>\:0\end{array}\;\;\Rightarrow\;\;\begin{array}  {cc}k < \frac{18}{7} \\ k > \frac{2}{5}\end{array}

    And there are only two solutions: . \begin{array}{cc}k=1: & (11,3) \\ k=2: & (4,8)\end{array}

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    Of course, this is a long and ponderous method . . . I warned you.

    There are far better methods (such as the Euclidean Algorithm).

    I wanted to show you that, if desperate,
    . . you can baby-talk your way through these problems.

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  7. #7
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    x=228+(35/5)t=228+7t
    y=-152-(25/5)t=-152-5t
    ..
    We want positive solutions.
    x,y>0
    Thus,
    228+7t>0
    -152-5t>0
    Solve,
    t>-32.5.....
    t<-30.4
    Thus,
    -32.5<t<-30.4
    We need integers.
    Those are,
    t=-32,-31
    Put them into the x,y equations to get the two possible solutions.
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