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L.C.M.一題!!!!!!!!!!

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已知兩正整數之和為1092，其L.C.M.為3528。 求二數。 要過程!!!!!!!

最佳解答:

x + y = 1092 x and y have a H.C.F. of p so that x = ap and y = bp. L.C.M. of x and y will be abp = 3528. ap + bp = 1092 p (a + b) = 1092 also a factor of p Therefore, p is H.C.F. of 1092 and 3528 as well. 3528 / 1092 = 3 ... 252 1092 / 252 = 4 ...84 252/84 = 3...0 0/84 = 0 H.C.F. of 3528 and 1092 is 84. These 2 numbers must be a factor of 84. Let those 2 numbers are 84a and 1092 - 84a. L.C.M. of these 2 numbers are 84a (1092-84a) / 84 = 3528. a(1092-84a) = 3528 a(13-a)=42 a^2-13a+42=0 a=6 or a=7 That 2 numbers are 504 and 588.

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