From the consecutive integers -10 to 10 inclusive,

This topic has expert replies
Junior | Next Rank: 30 Posts
Posts: 18
Joined: Fri Nov 10, 2017 9:01 am
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

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 16207
Joined: Mon Dec 08, 2008 6:26 pm
Location: Vancouver, BC
Thanked: 5254 times
Followed by:1268 members
GMAT Score:770

by Brent@GMATPrepNow » Mon Nov 20, 2017 9:38 am
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
Brent Hanneson - Creator of GMATPrepNow.com
Image

Junior | Next Rank: 30 Posts
Posts: 18
Joined: Fri Nov 10, 2017 9:01 am

by kamalj » Tue Nov 21, 2017 9:37 am
Choose nineteen -10's and one 10
Is there any particular reason to select nineteen -10 and one 10? Can we select another number? If yes can you please give me an example.

GMAT/MBA Expert

User avatar
Elite Legendary Member
Posts: 10392
Joined: Sun Jun 23, 2013 6:38 pm
Location: Palo Alto, CA
Thanked: 2867 times
Followed by:511 members
GMAT Score:800

by [email protected] » Tue Nov 21, 2017 12:29 pm
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
Contact Rich at [email protected]
Image

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 7245
Joined: Sat Apr 25, 2015 10:56 am
Location: Los Angeles, CA
Thanked: 43 times
Followed by:29 members

by Scott@TargetTestPrep » Sat Oct 12, 2019 4:04 pm
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

Scott Woodbury-Stewart
Founder and CEO
[email protected]

Image

See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews

ImageImage