AMC10 2003 A

AMC10 2003 A · Q15

AMC10 2003 A · Q15. It mainly tests Inclusion–exclusion (basic), Probability (basic).

What is the probability that an integer in the set $\lbrace 1, 2, 3, \dots , 100\rbrace$ is divisible by 2 and not divisible by 3?

集合 $\lbrace 1, 2, 3, \dots , 100\rbrace$ 中的整数能被 2 整除且不能被 3 整除的概率是多少？

(A) $\frac{1}{6}$ $\frac{1}{6}$

(B) $\frac{33}{100}$ $\frac{33}{100}$

(D) $\frac{1}{2}$ $\frac{1}{2}$

(E) $\frac{18}{25}$ $\frac{18}{25}$

Answer

Correct choice: (C)

正确答案：（C）

Solution

(C) Of the $\frac{100}{2}=50$ integers that are divisible by $2$, there are $\left\lfloor\frac{100}{6}\right\rfloor=16$ that are divisible by both $2$ and $3$. So there are $50-16=34$ that are divisible by $2$ and not by $3$, and $\frac{34}{100}=\frac{17}{50}$.

（C）在能被 $2$ 整除的 $\frac{100}{2}=50$ 个整数中，有 $\left\lfloor\frac{100}{6}\right\rfloor=16$ 个同时能被 $2$ 和 $3$ 整除。因此，能被 $2$ 整除但不能被 $3$ 整除的有 $50-16=34$ 个，并且 $\frac{34}{100}=\frac{17}{50}$。

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