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Ken 發問於 科學及數學數學 · 12 月前

It is given that 4p+q=9,where p, q are positive integers. Write down all possible values of p。請問是怎麼解答的?

2 個解答

評分
  • 12 月前
    最愛解答

    P=1, q=5

    P=2 ,q=1

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  • 匿名
    12 月前

    4p+q = 9 --- (1)

    Given that p, q are positive integers,

    p>0 --- (2)

    q>0 --- (3)

    4p+q = 9

    q = 9-4p --- (4)

    Sub (4) into (3),

    q > 0

    9-4p > 0

    4p < 9

    p < 9/4 (or 2.25)

    Since p is a positive integer, p = 1 or 2

    Sub p = 1 into (1),

    4(1)+q = 9

    q = 5

    Sub p = 2 into (1),

    4(2)+q = 9

    q = 1

    Therefore

    p = 1, q = 5

    p = 2, q = 1

    (另外看你要不要把 p>0也做一遍,會得出 q < 9,即 q = 1, 2, 3, 4, 5, 6, 7, 8,

    但 q = 2, 3, 4, 6, 7, 8 都需要被reject掉, 因為當q = 2, 3, 4, 6, 7, 8 sub into (1),得出來的 p 會是非Integer)

    完整一點的話最好寫,

    可以把When q = 2, 3, 4, 6, 7, 8 得出來的那堆直接reject (即 q = 1, 2 (rej.), 3 (rej.) ....)

    不過如果這題佔分數不多,就不太需要寫 (看你情況)

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