2003 AMC 8 第 19 题

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所有题目均经美国数学协会(MAA)官方合法授权使用。

19.

1000100020002000 之间有多少个整数同时以 15,2015, 202525 为因数?

How many integers between 10001000 and 20002000 have all three of the numbers 15,20,15, 20, and 2525 as factors?

11

22

33

44

55

答案:C
知识点:最小公倍数区间内整数计数
难度评级:1430
解答:

如果一个数 xx 以这三个数为因数,那么它们的最小公倍数也必须整除 xx

这些数的质因数分解为 15=35, 15 = 3 \cdot 5, 20=225, 20 = 2^2 \cdot 5, 25=52. 25 = 5^2.

因此最小公倍数为 22352=300. 2^2 \cdot 3 \cdot 5^2 = 300.

300300 的倍数中,位于 1000100020002000 之间的是 1200,15001200, 150018001800

所以正确答案是 C

If a number xx has these three numbers as factors, then their least common multiple must also divide x.x.

These numbers have the following prime factorizations: 15=35, 15 = 3 \cdot 5,20=225, 20 = 2^2 \cdot 5,25=52. 25 = 5^2.

From these values, we get that the least common multiple is 22352=300. 2^2 \cdot 3 \cdot 5^2 = 300.

Therefore, the multiples of 300300 between 10001000 and 20002000 are 1200,1500,1200, 1500, and 1800.1800.

Thus, C is the correct answer.

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