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

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08 Jul 2012, 17:29

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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

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 \(10^1*(-10)^{19}=-10^{20}\).

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

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08 Jul 2012, 17:53

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 \(10^1*(-10)^{19}=-10^{20}\) .

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)?

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 \(10^1*(-10)^{19}=-10^{20}\) .

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 \(10^1*(10)^{-19}=-10^{20}\).
_________________

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

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08 Jul 2012, 18:35

Thanks for pointing that out!!! Silly me. But i'm still not sure, why the other answer choices are eliminated and why E is the least possible answer?? >_<

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).
_________________

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 \(- (10)^{20}\)? You need to select -1 and twenty 10s but you cannot select 21 numbers. You cannot have the product as negative \(10^{20}\). Hence (E) is not possible.

E is possible if you select 10 odd number of times and -10 the remaining number of times. For example:

\(10^1*(-10)^{19}=-10^{20}\) or \(10^3*(-10)^{17}=-10^{20}\) or \(10^5*(-10)^{15}=-10^{20}\)... or \(10^{19}*(-10)^{1}=-10^{20}\).

But i'm still not sure, why the other answer choices are eliminated and why E is the least possible answer?? >_<

Actually, the question means "smallest possible value of the product"

\(-10^{20}\) is the smallest value of the given 5 options. Since it is possible, so it is the answer. I guess you were thinking 'least possible' (as was I). The wording is definitely misleading.
_________________

But i'm still not sure, why the other answer choices are eliminated and why E is the least possible answer?? >_<

Actually, the question means "smallest possible value of the product"

\(-10^{20}\) is the smallest value of the given 5 options. Since it is possible, so it is the answer. I guess you were thinking 'least possible' (as was I). The wording is definitely misleading.

Yes, the question means "what is the smallest possible value of the product of the 20 integers?"
_________________

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

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24 Jul 2012, 21:59

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i missed out the bracket for -10^20. I was assuming that value to be positive as a negative number raised to positive power is positive. Nice problem![/edit]
_________________

I've failed over and over and over again in my life and that is why I succeed--Michael Jordan Kudos drives a person to better himself every single time. So Pls give it generously Wont give up till i hit a 700+

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 20 integers.

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

The smallest possible value of the product will be negative (since it has to be smallest) with highest absolute value i.e. the number should be as left on the number line as possible. Out of all the options, (E) is the smallest. Let's see if it is possible. If you select all 10s and only one -10, you will get \(-10*(10)^{19}\) which is the same as \(-(10)^{20}\). So it is possible to get the product given in (E).

Answer must be (E)

Notice that options (A) and (B) are positive.
_________________

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

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20 Jan 2014, 02:01

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Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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

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25 Aug 2014, 11:05

This question shows how important it is to carefully look at the symbols. One might be very easily deceived in thanking the question is actually saying --> (-10)^20.

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

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20 Jan 2015, 05:26

Hi,

I also chose the right answer quite quickly. However, I didn't use powers and I am not sure why we used powers.

So what I though it that we need an odd number of negative values to have a negative value. To have the greatest negative value we need -10 (the higest possible value), the greatest number of odd times; so in out case 19. We still have to use one number, so 10 would be the final choise, as it would make the negative 19(-10) 10 times greater.

However, for me it is: -10(19) * 10 (1) instead of 10^1 * -10^19 = -10^20.

I understand the 10^1, but -10(19) is not -10^19. This is where I am lost...

I also chose the right answer quite quickly. However, I didn't use powers and I am not sure why we used powers.

So what I though it that we need an odd number of negative values to have a negative value. To have the greatest negative value we need -10 (the higest possible value), the greatest number of odd times; so in out case 19. We still have to use one number, so 10 would be the final choise, as it would make the negative 19(-10) 10 times greater.

However, for me it is: -10(19) * 10 (1) instead of 10^1 * -10^19 = -10^20.

I understand the 10^1, but -10(19) is not -10^19. This is where I am lost...

So, why the powers?

You are given that you need to find the product of all 20 numbers. When you multiply -10 by -10, you get \((-10)^2\), not \(-10*2\) (which is -10 + (-10))

Just like you multiplied the last 10 you picked, you have to multiply all previous 10s too. So multiply -10, 19 times to get \(-10*-10*-10...*-10 = (-10)^{19}\)
_________________

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

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27 Mar 2016, 08:23

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 \(-(10)^{20}\)

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,

we have to choose all 10s and ensure the combination is such that the sign becomes -ive.. \(10^{19}*(-10)..\) or \((-10)^{19}*10= -10^{20}\)

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