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

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

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30 Jun 2019, 12:27
Bunuel wrote:
catty2004 wrote:

I'm not quiet sure i understand, why choose only -10 and 10, there are other integers included in the set of -10 to 10. Also, isn't 10^1*10^(-19) = 10^(-18)?

We choose -10 and 10 because this way we get the least product (notice that -10^(20) is the smallest number among answer choices).

Also it's $$10^1*(-10)^{19}=-10^{20}$$ not $$10^1*(10)^{-19}=-10^{20}$$.

Hi Bunuel
Is it possible to prove the highlighted part?

Easy: $$10^1*(-10)^{19}=-1*10^1*10^{19}=-1*10^{1+19}=-10^{20}$$
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From the consecutive integers -10 to 10 inclusive, 20  [#permalink]

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30 Jun 2019, 12:34
Bunuel wrote:
Bunuel wrote:

We choose -10 and 10 because this way we get the least product (notice that -10^(20) is the smallest number among answer choices).

Also it's $$10^1*(-10)^{19}=-10^{20}$$ not $$10^1*(10)^{-19}=-10^{20}$$.

Hi Bunuel
Is it possible to prove the highlighted part?

Easy: $$10^1*(-10)^{19}=-1*10^1*10^{19}=-1*10^{1+19}=-10^{20}$$

Hi
I am talking about the highlighted part..The black circle is POSITIVE and the red circle is NEGATIVE. How they are equal each other?
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Re: From the consecutive integers -10 to 10 inclusive, 20  [#permalink]

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30 Jun 2019, 12:47
Hi
I am talking about the highlighted part..The black circle is POSITIVE and the red circle is NEGATIVE. How they are equal each other?

I'm saying there that $$10^1*(-10)^{19}=-10^{20}$$ NOT $$10^1*(10)^{-19}=-10^{20}$$ (which was written in the user's post).
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From the consecutive integers -10 to 10 inclusive, 20  [#permalink]

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30 Jun 2019, 12:54
Bunuel wrote:
Hi
I am talking about the highlighted part..The black circle is POSITIVE and the red circle is NEGATIVE. How they are equal each other?

I'm saying there that $$10^1*(-10)^{19}=-10^{20}$$ NOT $$10^1*(10)^{-19}=-10^{20}$$ (which was written in the user's post).

User wrote 10^(-18). And you wrote -10^20. I thought that you're trying to convince that we can have -10^20 by SEVERAL ways and the highlighted part (my highlighted part that I made) is ONE.
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Re: From the consecutive integers -10 to 10 inclusive, 20  [#permalink]

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30 Jun 2019, 13:05
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

Bunuel
Here are 21 numbers in the list (including 0). If we consider 0, then the product will be 0 (c). Why don't we consider 0 as a number?
Thanks__
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Re: From the consecutive integers -10 to 10 inclusive, 20  [#permalink]

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30 Jun 2019, 13:08
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

Bunuel
Here are 21 numbers in the list (including 0). If we consider 0, then the product will be 0 (c). Why don't we consider 0 as a number?
Thanks__

We can get 0, yes. But the question asks about the least possible product, and the least product is less than 0, it's negative: –(10)^20.
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Re: From the consecutive integers -10 to 10 inclusive, 20  [#permalink]

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01 Jul 2019, 06:00
Thanks for your nice explanation. What if the word "inclusive" is removed from the question stem?
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

The consecutive integers from -10 to 10 are: -10, -9, -8, -7 ... -1, 0, 1, ... 9, 10

You have to select 20 integers with repetition.

So you can select all 10. The product will be $$10^{20}$$.
or
You can select all -10. The product will be $$(-10)^{20}$$ (which is essentially same as $$10^{20}$$)
or
You can select ten 1s and ten -10s. The product will be $$(-10)^{10}$$
or
You can select 0 and any other 19 numbers. The product will be 0.
or
You can select -1 and nineteen 10s. The product will be $$-(10)^{19}$$.

But how will you get $$- (10)^{20}$$?

Editing here itself - I missed out the case in which you can get $$- (10)^{20}$$ (by multiplying odd number of -10 with rest of the 10s). Possibly because I mis-read the question.
'Least possible value of the product' to me implied the product that is not possible. Since all options a to d were possible, I overlooked that option (E) is possible too.

Actually the question should read 'the smallest possible value of the product' to avoid such confusion. Since -10^20 is the smallest value that can be obtained (greatest magnitude but negative), the answer is (E).

The word "inclusive" just makes us include 10 and -10 in the list of numbers. If these numbers are excluded, the smallest number would be $$-9 * 9^{19} = - 9^{20}$$
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From the consecutive integers -10 to 10 inclusive, 20  [#permalink]

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09 Aug 2019, 11:04
I felt something fishy in this question when I looked at option A and then at option E. I thought, how could the two be different from each other?
Quote:

A (−10)^20(−10)^20

E −(10)^20

But, you realize that you can take -10^19 and then simply multiply it again with 10 so that you get -10^19 (which is negative itself), times ten, making it the highest magnitude among the options, but negative, hence making it the least possible product.

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From the consecutive integers -10 to 10 inclusive, 20   [#permalink] 09 Aug 2019, 11:04

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