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

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

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28 Mar 2016, 02:01
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.
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Re: From the consecutive integers -10 to 10 inclusive, 20  [#permalink]

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04 May 2016, 10:49
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..

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Re: From the consecutive integers -10 and 10, inclusive, 20 integers are r  [#permalink]

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08 Dec 2016, 10:33
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: $$10^{19}*(-10) = (-1)*10^{20} = -(10)^{20}$$

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Re: From the consecutive integers -10 and 10, inclusive, 20 integers are r  [#permalink]

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08 Dec 2016, 10:44
(-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)
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Re: From the consecutive integers -10 to 10 inclusive, 20  [#permalink]

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18 Dec 2016, 07:00
i cannot even approach this kind of exercises, can someone elaborate? i would be grateful
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Re: From the consecutive integers -10 to 10 inclusive, 20  [#permalink]

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18 Dec 2016, 23:03
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Re: From the consecutive integers -10 to 10 inclusive, 20  [#permalink]

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19 Dec 2016, 04:18
thank you a lot bunuel
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Re: From the consecutive integers -10 to 10 inclusive, 20  [#permalink]

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25 Aug 2017, 10:46
See attached pic
Attachments

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Re: From the consecutive integers -10 to 10 inclusive, 20  [#permalink]

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15 Dec 2017, 00:48
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
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From the consecutive integers -10 to 10 inclusive, 20  [#permalink]

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15 Dec 2017, 05:45
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?
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Re: From the consecutive integers -10 to 10 inclusive, 20  [#permalink]

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05 Jan 2018, 08:24
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!
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Re: From the consecutive integers -10 to 10 inclusive, 20  [#permalink]

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05 Jan 2018, 08:31
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.
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Re: From the consecutive integers -10 to 10 inclusive, 20  [#permalink]

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05 Jan 2018, 08:42
$$-2^2 = -(2*2) = -4$$.
$$(-2)^2 = (-2)*(-2) = 4$$.

$$-10^{20}$$ 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 ?
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From the consecutive integers -10 to 10 inclusive, 20  [#permalink]

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05 Jan 2018, 08:46
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 $$-10^{20}=-(10*10*...*10*10*)$$
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From the consecutive integers -10 to 10 inclusive, 20 &nbs [#permalink] 05 Jan 2018, 08:46

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