求问OG13 PROBLEM SOLVING 196题

sissielizabeth

求教og 13 problem solving 196 题 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！！！

唐驿

这题是问从-10到10中随机抽取可重复的20个数求乘积，最小的乘积可能是多少。最小的肯定是个负数，且其绝对值最大，所以应该抽取的不是-10就是+10。然后答案就是-(10)^20 了，即抽取奇数个-10相乘，再乘上相应个+10，如11个-10，9个+10相乘。
不知道解释明白没有。。。好像应该是这样吧。。。