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數學知識交流---取出正整數(Part 2b)

發問:

從 1 , 2 , 3 , ... , 999 , 1000 共 1000 個正整數中,最多能取出多少個正整數,使得對於取出來的正整數中任意四個數 a , b , c , d ( a < b < c < d ),都有 abc ≠ d。

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Consider abc > a^3 then if a > 1000^(1/3) = 10, abc > d. Therefore (11~1000) all fits the condition. Consider 10: 10*b*c > 10^3 > 1000, therefore 10 fits the condition. Consider 9; 9*10*11 = 990, therefore 9 does not fit the condition. Now we prove that {10,11,...1000} is a global maximum: When we choose a number x < 10, then the rest of the number must fulfill the condition that: (I) k > 10, OR (II) k > (1000/x)^(1/2) > 10 which is a looser bounding. Therefore setting x > 10 is the best condition. Answer = 991 (10 to 1000).

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