kamalj 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

OAE

Choose nineteen -10's and one 10
So, the product = [(-10)^19][10]
Notice that [(-10)^19] is NEGATIVE, which means [(-10)^19][10] is also NEGATIVE.

So, [(-10)^19][10] = -[(10)^19][10]
= -(10)^20
= E

Cheers,
Brent

Hi kamalj,

The question asks us for the LEAST possible product, so we need a product that is as 'negative' as possible (thus, we need to think about the 'extreme' values: a mix of +10s and -10s). There are actually a number of different ways to get that same product (but the number of -10s would have to be an ODD number). For example:

One -10 and nineteen +10s
Three -10s and seventeen +10s
Five -10s and fifteen +10s
Etc.

GMAT assassins aren't born, they're made,
Rich

kamalj 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

OAE

This problem is testing our knowledge of the multiplication rules for positive and negative numbers. Remember that when we multiply an even number of negative numbers together, the result is positive, and when we multiply an odd number of negative numbers together, the result is negative.

Because we are selecting 20 numbers from the list, we want to start by selecting the smallest 19 numbers and multiplying those together. In our list, the smallest number we can select is -10. So, we have:

(-10)^19 (Note that this product will be negative.)

Since we need to select a total of 20 numbers, we must select one additional number from the list. However, since the final product must be as small as possible, we want the final number we select to be the largest positive value in our list. The largest positive value in our list is 10. So, the product of our 20 integers is:

(-10)^19 x 10 (Note that this product will still be negative.)

This does not look identical to any of our answer choices. However, notice that (-10)^19 can be rewritten as -(10)^19, so:

(-10)^19 x 10 = -(10)^19 x (10)^1 = -(10)^20

Answer: E