1995 AMC 12 第 29 题

先试着解答 1995 AMC 12 第 29 题,然后核对你的答案与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 1995 AMC 12 解答,或核对答案。

所有题目均经美国数学协会(MAA)官方合法授权使用。

29.

有多少个由三个正整数组成的集合 {a,b,c}\{a,b,c\} 满足 a⋅b⋅c=2310a\cdot b\cdot c=2310?
For how many three-element sets of positive integers {a,b,c}\{a,b,c\} is it true that a⋅b⋅c=2310?a\cdot b\cdot c=2310?
3232
3636
4040
4343
4545
答案:C
知识点:质因数分解set partitions
小提示:
将 23102310 分解为互不相同的质因数,并把每个质因数分配给一个因数
Factor 23102310 into distinct primes and assign each prime to one of the factors
大提示:
分别统计一个因数为 11 以及三个因数都大于 11 的情形
Count separately the cases in which one factor is 11 and in which all three factors exceed 11
解答:
因为 2310=2⋅3⋅5⋅7⋅112310=2\cdot3\cdot5\cdot7\cdot11,每个质因数恰好属于三个因数中的一个。若三个因数都大于 11,它们对应的无序质因数组就是把五个对象分成三个非空组,其数量为 S(5,3)=35−3⋅25+36=25。 S(5,3)=\frac{3^5-3\cdot2^5+3}{6}=25\text{。}若有一个因数为 11,则把这些质因数分成两个非空组,得到 S(5,2)=24−1=15S(5,2)=2^4-1=15。由于互不相交的质因数集合不同,所得因数也互不相同。因此总数为 25+15=4025+15=40,正确答案是 C。
Since 2310=2⋅3⋅5⋅7⋅11,2310=2\cdot3\cdot5\cdot7\cdot11, each prime belongs to exactly one of the three factors. If all factors exceed 1,1, their unordered prime groups form a partition of five objects into three nonempty blocks, counted by S(5,3)=35−3⋅25+36=25. S(5,3)=\frac{3^5-3\cdot2^5+3}{6}=25. If one factor is 1,1, the primes are partitioned into two nonempty blocks, giving S(5,2)=24−1=15.S(5,2)=2^4-1=15. The factors are distinct because their disjoint prime sets differ. Thus the total is 25+15=40,25+15=40, and the correct answer is C.
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