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1. If b is a positive integer than 200, then how many integer pairs (a,b) satisfy the equation a/b= 2/9

  • math -

    I assume you meant:
    If b is a positive integer < 200

    a/b = 2/9
    9a = 2b
    a = 2b/9

    So b must be a multiple of 9 to have a as an integer
    the smallest value of b is 9
    and the largest value of b is 198
    so we have:
    9 18 27 .... 180 189 198

    to find out how many terms are in this arithmetic sequence ...
    term(n) = a + (n-1)d
    198 = 9 + (n-1)(9)
    189 = 9n - 9
    198 = 9n
    n = 22

    there are 22 such ordered pairs

    check:

    the pairs are (2,9) , (4,18) , (6,27) , ... (42, 189) , (44, 198)
    in each case a/b = 2/9
    e.g.1 :
    27/6 = 2/9
    189/42 = 2/9
    ....
    etc

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