求教og 13 problem solving 196 题

sissielizabeth · 发表于 2012-8-31 23:56:52

From the consecutive integers -10 to 10, inclusive, 20 integers are randomly chosen with repitions allowed.What is the least possible value of the product of the 20 integers?
A (-10)^20
B (-10)^10
C 0
D -(10)^19
E -(10)^20
Answer:
If -10 is chosen an odd number of times and 10 is chosen the remaining number of times (for example, choose -10 once and choose 10 nineteen times, or choose -10 three times and choose 10 seventeen times), then the product of the 20 chosen numbers will be
-(10)^20.Note that -(10)^20 is less than -(10)^19,the only other negative value among the answer choices.
the correct answer is E

我看不懂这个答案啊！！不是问的是从-10 到10 直间的数随机抽20个然后乘吗，为什么只看10和-10.。.。而且为什么还分什么奇偶次数！！
等待大神出现！！！

sissielizabeth · 发表于 2012-9-1 09:43:58

顶上去

zzc880315 · 发表于 2012-9-1 10:08:07

-10到10之间随即取20个相乘，乘积可正可负，所以最小值肯定是负数

取奇数个-10就是保证最后的值是负的

sissielizabeth · 发表于 2012-9-1 10:15:07

明白了！！Least possible vlaue 是最小的可能值我理解成最小可能的值

sissielizabeth · 发表于 2012-9-1 10:28:29

谢谢！！

求教og 13 problem solving 196 题

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