Problem Solving

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






Answer: E

You know you can get an odd number of negative numbers, so the product can be negative. Eliminate A, B, and C. Then, you have to decide between D and E. If you have an odd number of -10's and and odd number of 10's you have -(10)^20.

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?

To get the smallest value, we need to
A. Maximize the absolute value
B. Stick a negative sign in front of it

Now it becomes easy.
Since 10 and -10 are the largest and smallest numbers,
The largest absolute value is 10*10*10.....10 (20 times)= 10^20
Hence smallest value is -(10^20)
( if we select an odd no. of -10s, and odd no. of 10s, we get the answer above!
Hence E

To solidify the concept try this question:

If 10 integers are randomly selected from the following set S, when each integer can be selected any number of times, what is the smallest possible value of the product of the selected integers?
Set S: {-1000, -100, -10, 1, 100}

A: -(10^26)
B: -(10^27)
C: -(10^28)
D: -(10^29)
E: -(10^30)

OA later

eagleeye wrote:

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?

To get the smallest value, we need to
A. Maximize the absolute value
B. Stick a negative sign in front of it

Now it becomes easy.
Since 10 and -10 are the largest and smallest numbers,
The largest absolute value is 10*10*10.....10 (20 times)= 10^20
Hence smallest value is -(10^20)
( if we select an odd no. of -10s, and odd no. of 10s, we get the answer above!
Hence E

To solidify the concept try this question:

If 10 integers are randomly selected from the following set S, when each integer can be selected any number of times, what is the smallest possible value of the product of the selected integers?
Set S: {-1000, -100, -10, 1, 100}

A: -(10^26)
B: -(10^27)
C: -(10^28)
D: -(10^29)
E: -(10^30)

OA later

I have done it this way.
In the set S the least number is -1000. Now if we select it 10 times then it would be (-1000)^10. But this would become a positive number because (-1)^10 (1000)^10 = (1000)^10

So select -1000 9 times and the other number should be the largest positive number in the set. So it would be (-1000)^9 (100) = -(10^29)
Answer : D

coolhabhi wrote:

eagleeye wrote:

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?

To get the smallest value, we need to
A. Maximize the absolute value
B. Stick a negative sign in front of it

Now it becomes easy.
Since 10 and -10 are the largest and smallest numbers,
The largest absolute value is 10*10*10.....10 (20 times)= 10^20
Hence smallest value is -(10^20)
( if we select an odd no. of -10s, and odd no. of 10s, we get the answer above!
Hence E

To solidify the concept try this question:

If 10 integers are randomly selected from the following set S, when each integer can be selected any number of times, what is the smallest possible value of the product of the selected integers?
Set S: {-1000, -100, -10, 1, 100}

A: -(10^26)
B: -(10^27)
C: -(10^28)
D: -(10^29)
E: -(10^30)

OA later

I have done it this way.
In the set S the least number is -1000. Now if we select it 10 times then it would be (-1000)^10. But this would become a positive number because (-1)^10 (1000)^10 = (1000)^10

So select -1000 9 times and the other number should be the largest positive number in the set. So it would be (-1000)^9 (100) = -(10^29)
Answer : D

Well done coolhabhi . You got the right approach and answer