Thanks for your nice explanation.
catty2004 wrote:
From the consecutive integers -10 to 10 inclusive, 20 integers are randomly chosen with repetitions 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
The consecutive integers from -10 to 10 are: -10, -9, -8, -7 ... -1, 0, 1, ... 9, 10
You have to select 20 integers with repetition.
So you can select all 10. The product will be \(10^{20}\).
or
You can select all -10. The product will be \((-10)^{20}\) (which is essentially same as \(10^{20}\))
or
You can select ten 1s and ten -10s. The product will be \((-10)^{10}\)
or
You can select 0 and any other 19 numbers. The product will be 0.
or
You can select -1 and nineteen 10s. The product will be \(-(10)^{19}\).
But how will you get \(- (10)^{20}\)?
Editing here itself - I missed out the case in which you can get \(- (10)^{20}\) (by multiplying odd number of -10 with rest of the 10s). Possibly because I mis-read the question.
'Least possible value of the product' to me implied the product that is not possible. Since all options a to d were possible, I overlooked that option (E) is possible too.
Actually the question should read 'the smallest possible value of the product' to avoid such confusion. Since -10^20 is the smallest value that can be obtained (greatest magnitude but negative), the answer is (E).