1959 AMC 12 第 42 题

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42.

给定三个正整数 aabbcc。它们的最大公因数为 DD,最小公倍数为 MM。下列哪两个陈述正确?

(1)(1) 乘积 MDMD 不可能小于 abcabc

(2)(2) 乘积 MDMD 不可能大于 abcabc

(3)(3) 当且仅当 aabbcc 都是质数时,MDMD 等于 abcabc

(4)(4) 当且仅当 aabbcc 两两互质时,MDMD 等于 abcabc。(即任意两个数都没有大于 11 的公因数。)

Given three positive integers a,a, b,b, and c.c. Their greatest common divisor is D;D; their least common multiple is M.M. Then, which two of the following statements are true?

(1)(1) The product MDMD cannot be less than abc.abc.

(2)(2) The product MDMD cannot be greater than abc.abc.

(3)(3) MDMD equals abcabc if and only if a,a, b,b, cc are each prime.

(4)(4) MDMD equals abcabc if and only if a,a, b,b, cc are relatively prime in pairs. (This means: no two have a common factor greater than 1.1.)

1122

1,1, 22

1133

1,1, 33

1144

1,1, 44

2233

2,2, 33

2244

2,2, 44

答案:E
知识点:最大公约数最小公倍数质因数分解
难度评级:1790
小提示:

对任一质数,将它在 a,b,ca,b,c 中的指数按 uvwu\le v\le w 排列

For one prime, order its exponents in a,b,ca,b,c as uvwu\le v\le w

大提示:

比较它在 MDMD 中的指数 u+wu+w 与在 abcabc 中的指数 u+v+wu+v+w

Compare the exponent u+wu+w in MDMD with the exponent u+v+wu+v+w in abcabc

解答:

对任一质数,设它在 a,b,ca,b,c 中的指数为 uvwu\le v\le w。它在 MDMD 中的指数为 u+wu+w,而在 abcabc 中的指数为 u+v+wu+v+w。因此 MDabcMD\le abc,证明了陈述 (2)(2)。对每个质数都恰有 v=0v=0 时等号成立,这意味着没有任何质数同时整除 a,b,ca,b,c 中的两个数。这正是两两互质,证明了陈述 (4)(4)

因此,正确答案是 E

For any prime, let its exponents in a,b,ca,b,c be uvw.u\le v\le w. Its exponent in MDMD is u+w,u+w, while its exponent in abcabc is u+v+w.u+v+w. Thus MDabc,MD\le abc, proving statement (2).(2). Equality holds exactly when v=0v=0 for every prime, meaning no prime divides two of a,b,c.a,b,c. That is precisely pairwise relative primality, proving statement (4).(4).

Therefore, the correct answer is E.

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