GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

 It is currently 25 May 2020, 04:37 ### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # From the consecutive integers -10 to 10 inclusive, 20 integers are ran

Author Message
TAGS:

### Hide Tags

Math Expert V
Joined: 02 Sep 2009
Posts: 64096
Re: From the consecutive integers -10 to 10 inclusive, 20 integers are ran  [#permalink]

### Show Tags

Mo2men wrote:
Bunuel wrote:
mbaspire 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 $$-(10)^{20}$$

Hi Bunuel,

I'm confused. I have referred to gmat club math book about multiply 2 number with exponents. the base should be the same to allow sum of the exponent while in this question the base is 10 & -10. So we can't sum 10^10 * (-10)^19. Do I miss something?

Thanks

Notice that (-10)^19 = (-1)*(10)^19. This way you'll get the same base.

Hope it's clear.
_________________
Target Test Prep Representative V
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 10543
Location: United States (CA)
Re: From the consecutive integers -10 to 10 inclusive, 20 integers are ran  [#permalink]

### Show Tags

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

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.

_________________

# Scott Woodbury-Stewart

Founder and CEO

Scott@TargetTestPrep.com

See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews

If you find one of my posts helpful, please take a moment to click on the "Kudos" button.

Board of Directors D
Status: QA & VA Forum Moderator
Joined: 11 Jun 2011
Posts: 4989
Location: India
GPA: 3.5
Re: From the consecutive integers -10 to 10 inclusive, 20 integers are ran  [#permalink]

### Show Tags

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

Here is something interesting for those who might have hard time with this question..

Think of the question in smaller terms

Quote:
From -10 to 10 ; select 4 numbers (repetition allowed) having the least possible product

Possible numbers are

{ -10 , -9, -8, -7, -6, -5, -4, -3, -2, -1 , 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

Now you can select any number right ? But what is the best way to select least possible number ?

Take 3 negative numbers (repetition allowed)

$${-10}^3$$ = -1000

Can you make any smaller number than this ? Don't stress, you can't ...

Now you have product of 3 negative numbers , multiplying with another negative number will result in a positive number, so that is not our objective...

So select a positive number . But which number to choose 1 or 10, lets see..

$${-10}^3$$ x 1 => $$-1000$$

$${-10}^3$$ x 10 => $$-10000$$

Which is smaller -1000 or -10000 ?

It's definitely -10000....

The similar logic applies to this problem..

Hope it helps someone, having difficulty understanding this problem..

_________________
Thanks and Regards

Abhishek....

PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS

How to use Search Function in GMAT Club | Rules for Posting in QA forum | Writing Mathematical Formulas |Rules for Posting in VA forum | Request Expert's Reply ( VA Forum Only )
Senior Manager  B
Joined: 13 Oct 2016
Posts: 352
GPA: 3.98
Re: From the consecutive integers -10 to 10 inclusive, 20 integers are ran  [#permalink]

### Show Tags

In order to minimize the product we can choose 19 biggest positive integers and then multiply their product by the smallest integer. That is |X| should be max if repetition is allowed.

We have: $$10^{19}*(-10) = (-1)*10^{20} = -(10)^{20}$$

Manager  B
Joined: 03 Oct 2013
Posts: 76
Re: From the consecutive integers -10 to 10 inclusive, 20 integers are ran  [#permalink]

### Show Tags

1
(-10)^20 is a positive number. So let's limit to (-10)^19.

now the last number should still keep this negative and minimize it further. for example, 1x (-10)^19>2x(-10)^19.....so on..

Therefore the minimum this can be is 10x(-10)^19 = -1x(10^20)
Senior Manager  P
Joined: 29 Jun 2017
Posts: 411
GPA: 4
WE: Engineering (Transportation)
Re: From the consecutive integers -10 to 10 inclusive, 20 integers are ran  [#permalink]

### Show Tags

See attached pic
Attachments IMG_9182.JPG [ 1.74 MiB | Viewed 1718 times ]

Veritas Prep GMAT Instructor V
Joined: 16 Oct 2010
Posts: 10441
Location: Pune, India
Re: From the consecutive integers -10 to 10 inclusive, 20 integers are ran  [#permalink]

### Show Tags

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

Here is the video solution to this interesting number properties question: https://www.veritasprep.com/gmat-soluti ... olving_208
_________________
Karishma
Veritas Prep GMAT Instructor

Intern  B
Joined: 27 Jul 2016
Posts: 3
Re: From the consecutive integers -10 to 10 inclusive, 20 integers are ran  [#permalink]

### Show Tags

I have used a totally a different aproach that came to my mind, please tell me if my logic is correct. I got the answer right

We have number of consecutive integers from -10 to 10 INCLUSIVE. This means that there 21 integers, or 21 picks we can make.
Pick 10 or -10 and rise it to power of 20, since we can pick one number for 20 time and then multiply the positive number you get by (-1) which is the 21st final pick you can make, so you will get:

+-10ˆ20 * (-1) = - (10)ˆ20

I arrived to the correct answer. Is my approach/thinking correct?
VP  D
Joined: 09 Mar 2016
Posts: 1235
Re: From the consecutive integers -10 to 10 inclusive, 20 integers are ran  [#permalink]

### Show Tags

Bunuel 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?

A. (-10)^20
B. (-10)^10
C. 0
D. –(10)^19
E. –(10)^20

Select 10 odd number of times and -10 the remaining number of times, for example select 10 once and -10 nineteen times, then the product will be $$10^1*(-10)^{19}=-10^{20}$$.

Hello Bunuel, hope you have a fantastc day Can you explain pleaseeee,what is the difference between (-10)^20 and –(10)^20 ? yes I see minus sign – that’s why I don’t understand…. Also how can this number be the least number/ smallest one –(10)^20 if we multiply 10 TWENTY times by itself we get huge number... Thanks!
Math Expert V
Joined: 02 Sep 2009
Posts: 64096
Re: From the consecutive integers -10 to 10 inclusive, 20 integers are ran  [#permalink]

### Show Tags

1
1
dave13 wrote:
Bunuel 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?

A. (-10)^20
B. (-10)^10
C. 0
D. –(10)^19
E. –(10)^20

Select 10 odd number of times and -10 the remaining number of times, for example select 10 once and -10 nineteen times, then the product will be $$10^1*(-10)^{19}=-10^{20}$$.

Hello Bunuel, hope you have a fantastc day Can you explain pleaseeee,what is the difference between (-10)^20 and –(10)^20 ? yes I see minus sign – that’s why I don’t understand…. Also how can this number be the least number/ smallest one –(10)^20 if we multiply 10 TWENTY times by itself we get huge number... Thanks!

$$-2^2 = -(2*2) = -4$$.
$$(-2)^2 = (-2)*(-2) = 4$$.

$$-10^{20}$$ is -(huge number), so it's has very small value. The same way as say, -100 is less than -10.
_________________
VP  D
Joined: 09 Mar 2016
Posts: 1235
Re: From the consecutive integers -10 to 10 inclusive, 20 integers are ran  [#permalink]

### Show Tags

$$-2^2 = -(2*2) = -4$$.
$$(-2)^2 = (-2)*(-2) = 4$$.

$$-10^{20}$$ is -(huge number), so it's has very small value. The same way as say, -100 is less than -10.[/quote]

Bunuel thanks! one question $$-10^{20}$$ -10 is raised to even power which means that negative sign disappears, and number becomes positive during multiplication by itself no ?
Math Expert V
Joined: 02 Sep 2009
Posts: 64096
Re: From the consecutive integers -10 to 10 inclusive, 20 integers are ran  [#permalink]

### Show Tags

dave13 wrote:
$$-2^2 = -(2*2) = -4$$.
$$(-2)^2 = (-2)*(-2) = 4$$.

$$-10^{20}$$ is -(huge number), so it's has very small value. The same way as say, -100 is less than -10.

Bunuel thanks! one question $$-10^{20}$$ -10 is raised to even power which means that negative sign disappears, and number becomes positive during multiplication by itself no ?

Consider the examples in my post (highlighted).

If it were $$(-10)^{20}$$ you'd be right but it's $$-10^{20}=-(10*10*...*10*10*)$$
_________________
Intern  B
Joined: 15 Sep 2018
Posts: 31
Re: From the consecutive integers -10 to 10 inclusive, 20 integers are ran  [#permalink]

### Show Tags

1
To get the smallest possible product, we need to consider the smallest number from this set: that would be $$-10$$.

We are allowed to have duplicate numbers, but we have to be careful about choosing this number 20 times.

If we did, then we will get:
$$(-10)^{20}$$

This ends up being a large, positive number. In order to get the smallest overall product, we need to think in terms of a negative solution.

By multiplying a large product by a negative number, we can find that negative product that we're looking for. There are two possibilities:

$$10 \times (-10)^{19}\\ 10^{19} \times (-10)$$

The smallest possible product is $$–(10^{20})$$.

Manager  P
Joined: 14 Apr 2017
Posts: 77
Location: Hungary
GMAT 1: 760 Q50 V42 WE: Education (Education)
Re: From the consecutive integers -10 to 10 inclusive, 20 integers are ran  [#permalink]

### Show Tags

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

Each of these randomly selected integers could be a negative number, a positive number, or zero, so their product could be negative. Their product has the least possible value if it is a negative number with the greatest possible absolute value.

First, we select the number with the greatest possible absolute value for each place, except one. Then, we must be careful to select the right number for the last place because we have to consider not only its absolute value but also its sign.

Although -10 and 10 have the same absolute value, we should choose 10s for the first 19 places because negative numbers always complicate things. Since $$10^{19}$$ is positive, we must select the negative number with the greatest possible absolute value for the last place in the product.

$$10^{19}(−10)=(−)10^{19}(10)=(−)10^{20}=−10^{20}$$

$$−10^{20}$$ is the same number as $$−(10)^{20}$$ in the correct answer choice.

If the given set of integers consisted of only non-negative or only non-positive numbers, then we would need different strategies.
_________________
My book with my solutions to all 230 PS questions in OG2018:
Zoltan's solutions to OG2018 PS
Intern  B
Joined: 18 Jul 2018
Posts: 30
Re: From the consecutive integers -10 to 10 inclusive, 20 integers are ran  [#permalink]

### Show Tags

I understand an odd number of neg. tens and the remainder of pos. tens will give -10^20. Why can't we interpret -10^20 as having 20 pos. tens and one -1( -1*10^20), therefore exceeding the number of randomly selected numbers allowed?
Intern  B
Joined: 18 Jul 2018
Posts: 30
Re: From the consecutive integers -10 to 10 inclusive, 20 integers are ran  [#permalink]

### Show Tags

If we're only allowed to select 20 integers, can't -(10)^20 be read as (-1)(10)^20 is therefore comprised of 21 integers? This would make D the correct answer. Why this reading incorrect? Re: From the consecutive integers -10 to 10 inclusive, 20 integers are ran   [#permalink] 19 Dec 2019, 21:43

Go to page   Previous    1   2   [ 36 posts ]

# From the consecutive integers -10 to 10 inclusive, 20 integers are ran  