AMC 8 2011 Problem 8
AMC8 2011 年第 8 题
基本计数原理★☆☆☆☆
Bag A has three chips labeled 1, 3, and 5. Bag B has three chips labeled 2, 4, and 6. If one chip is drawn from each bag, how many different values are possible for the sum of the two numbers on the chips?
口袋 A 中有标有 1，3，5 的三个筹码，口袋 B 中有标有 2，4，6 的三个筹码。若从每个口袋中随机抽出一个筹码，那么这 2 个筹码上的数字之和有多少种可能的不同值？
- A.4
- B.5
- C.6
- D.7
- E.9
正确答案 · Correct Answer
B
解析 · Solution
Solution 1
By adding a number from Bag A and a number from Bag B together, the values we can get are 3, 5, 7, 5, 7, 9, 7, 9, 11. Therefore the number of different values is 5.
The sum of an even number added to an odd number is always odd. The smallest possible sum is 3, and the largest possible sum is 11. The odd numbers in between can be achieved by replacing chips with within the same bag. Therefore, we can conclude that there are (11-3)/2(*)+1=5 possible sums.
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