AMC8 2009
AMC8 2009 · Q13
AMC8 2009 · Q13. It mainly tests Combinations, Divisibility & factors.
A three-digit integer contains one of each of the digits 1, 3 and 5. What is the probability that the integer is divisible by 5?
一个三位整数包含 1、3 和 5 各一个数字。该整数能被 5 整除的概率是多少？
(A)
\frac{1}{6}
\frac{1}{6}
(B)
\frac{1}{3}
\frac{1}{3}
(C)
\frac{1}{2}
\frac{1}{2}
(D)
\frac{2}{3}
\frac{2}{3}
(E)
\frac{5}{6}
\frac{5}{6}
Answer
Correct choice: (B)
正确答案：（B）
Solution
Answer (B): There are 6 three-digit numbers possible using the digits 1, 3 and 5 once each: 135, 153, 315, 351, 513 and 531. Because the numbers divisible by 5 end in 0 or 5, only 135 and 315 are divisible by 5. The probability that the three-digit number is divisible by 5 is $\frac{2}{6}=\frac{1}{3}$.
答案（B）：使用数字 1、3、5 各一次可以组成 6 个三位数：135、153、315、351、513、531。因为能被 5 整除的数末位为 0 或 5，所以只有 135 和 315 能被 5 整除。这个三位数能被 5 整除的概率是 $\frac{2}{6}=\frac{1}{3}$。
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