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Math Help - Pairs of (x,y)

  1. #1
    Junior Member Dragon's Avatar
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    Pairs of (x,y)

    How many pairs (x,y) of positive integers satisfiy 2x+7y=1000
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  2. #2
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    Quote Originally Posted by ginafara View Post
    The answer is all of the points x,y in Quadrant I, that lie on the line...

    y = (-2/7)x +1000

    when y = 0, x = 500 so all of the integers from 0 -> 500, since it is defined over its domain, it would be 500 pair (?)

    pretty sure that is your answer
    Hello,

    I don't want to pick at you but from

    2x + 7y = 1000 \ \Longleftrightarrow \ y=-\frac{2}{7}x + \frac{1000}{7}

    Since you were looking for integers for x and y you have to calculate how often 1000 is divided by 14. Thus I got 71 pairs.
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by Dragon View Post
    How many pairs (x,y) of positive integers satisfiy 2x+7y=1000
    <br />
y=\frac{1000-2x}{7}<br />

    so for y to be an integer 1000-2x must be a multiple of 7,
    but as this is even 500-x must be a multiple of 7.

    But as x ranges over the integers 1, .. , 500 (which is the range
    that x is constrained to) 500-x ranges over  500, .. , 1.

    Of the numbers  500, .. , 1 exactly  \lfloor 500/7 \rfloor=71 are divisible by 7,
    so there are 71 positive integer pairs (x,y) that satisfy the
    given equation.

    RonL
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  4. #4
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    Hello, Dragon!

    My approach is a variation of CaptainBlack's . . .


    How many pairs (x,y) of positive integers satisfy 2x+7y\:=\:1000

    We have: . y \:=\:\frac{1000-2x}{7}
    Since y is a positive integer, we have:. y \;=\;\frac{\text{even number }\leq 1000}{\text{divisible by 7}}

    There are: \left[\frac{1000}{7}\right] = 142 multiples-of-seven which are less than 1000

    . . and half of them are even.


    Therefore, there are 71 solutions.

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  5. #5
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    Quote Originally Posted by Dragon View Post
    How many pairs (x,y) of positive integers satisfiy 2x+7y=1000
    2x+7y=1000

    We can solve this using Bezout's Identity.

    Trivially we see that,
    2(-3)+7(1)=1
    Thus,
    2(-3000)+7(1000)=1000.

    Hence, all integer solutions are given by,
    x=-3000+7t
    y=1000-2t.

    We want positive solutions, meaning,
    x>0 \to -3000+7t >0 \to 7t > 3000 \to t> 428.5...
    y>0 \to 1000-2t >0 \to 2t <1000 \to t<500
    Since, t is an integer, we need to find all integers so that,
    428.5...<t<500
    We see that,
    429,430,431,....,499
    Are all these integers.
    In total we have 71 positive solutions.
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