AMC10 2001 A

AMC10 2001 A · Q25

AMC10 2001 A · Q25. It mainly tests Inclusion–exclusion (basic), Divisibility & factors.

How many positive integers not exceeding 2001 are multiples of 3 or 4 but not 5?

不超过 2001 的正整数中，能被 3 或 4 整除但不被 5 整除的有多少个？

(A) 768 768

(B) 801 801

(D) 1067 1067

(E) 1167 1167

Answer

Correct choice: (E)

正确答案：（E）

Solution

(B) For integers not exceeding 2001, there are $\lfloor 2001/3\rfloor = 667$ multiples of 3 and $\lfloor 2001/4\rfloor = 500$ multiples of 4. The total, 1167, counts the $\lfloor 2001/12\rfloor = 166$ multiples of 12 twice, so there are $1167-166=1001$ multiples of 3 or 4. From these we exclude the $\lfloor 2001/15\rfloor = 133$ multiples of 15 and the $\lfloor 2001/20\rfloor = 100$ multiples of 20, since these are multiples of 5. However, this excludes the $\lfloor 2001/60\rfloor = 33$ multiples of 60 twice, so we must re-include these. The number of integers satisfying the conditions is $1001-133-100+33=801$.

（B）对于不超过 2001 的整数，3 的倍数有 $\lfloor 2001/3\rfloor = 667$ 个，4 的倍数有 $\lfloor 2001/4\rfloor = 500$ 个。两者合计 1167，但其中 $\lfloor 2001/12\rfloor = 166$ 个 12 的倍数被重复计数了两次，所以 3 或 4 的倍数共有 $1167-166=1001$ 个。接着从中排除 $\lfloor 2001/15\rfloor = 133$ 个 15 的倍数和 $\lfloor 2001/20\rfloor = 100$ 个 20 的倍数，因为它们都是 5 的倍数。然而这样会把 $\lfloor 2001/60\rfloor = 33$ 个 60 的倍数重复排除两次，因此需要把它们加回来。满足条件的整数个数为 $1001-133-100+33=801$。

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