Problem Solving

DCS80 wrote:Hi,


#196 (OG 13)

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


Having issues grasping how D is distinguishable from E, and also how they are differentiated from A

thanks

Solution:

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 negative numbers we can 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.)

Because we need to select a total of 20 numbers, we must select one additional number, which must be positive, 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 = (-1 x 10)^19 = (-1)^19 x (10)^19 = - 1 x (10)^19 = -(10)^19, so

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

Answer: E

Hi All,

This question is built around Number Properties (Number Lines) and basic arithmetic (multiplication and exponents). Since we're multiplying 20 integers together and we want the SMALLEST PRODUCT possible, we need to consider HOW we can make that small product happen.

We're given the range of -10 to +10, inclusive AND we're allowed to repeat numbers. Since the range includes negative values, we can clearly end up with a negative product - it's just a matter of how 'negative' we can make it.

Conceptually, there are two ways to get the smallest number possible (meaning the FARTHEST LEFT on the number line):

(10^19)(-10)
[(-10)^19](10)

Each of these products will have the same end value...

Final Answer: E

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

DCS80 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

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

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

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 we can 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