AMC10 2011 A
AMC10 2011 A · Q13
AMC10 2011 A · Q13. It mainly tests Basic counting (rules of product/sum), Parity (odd/even).
How many even integers are there between 200 and 700 whose digits are all different and come from the set $\{1, 2, 5, 7, 8, 9\}$?
在200与700之间，有多少个偶数，其所有数位均不同且来自集合 $\{1, 2, 5, 7, 8, 9\}$？
(A)
12
12
(B)
20
20
(C)
72
72
(D)
120
120
(E)
200
200
Answer
Correct choice: (A)
正确答案：（A）
Solution
Because the numbers are even, they must end in either 2 or 8. If the last digit is 2, the first digit must be 5 and thus there are four choices remaining for the middle digit. If the last digit is 8, then there are two choices for the first digit, either 2 or 5, and for each choice there are four possibilities for the middle digit. The total number of choices is then $4 + 2 \cdot 4 = 12$.
因为这些数是偶数，末位必须是2或8。若末位是2，则首位必须是5，中间位有4种选择。若末位是8，则首位有2种选择（2或5），每种情况下中间位有4种可能。总选择数为 $4 + 2 \cdot 4 = 12$。
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