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Maths

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Intern
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Joined: 18 Aug 2017
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Maths  [#permalink]

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New post 19 Aug 2017, 03:32
From the consecutive integers -10 to 10 inclusive 20 integers are randomly chosen with repetition allowed. What is the least value of the product of the 20 integers?
A (-10)^20
B (-10)^10
C 0
D (-10)^19
E (-10)^20

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Re: Maths  [#permalink]

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New post 23 Aug 2017, 07:25
kartik0604 wrote:
From the consecutive integers -10 to 10 inclusive 20 integers are randomly chosen with repetition allowed. What is the least value of the product of the 20 integers?
A (-10)^20
B (-10)^10
C 0
D (-10)^19
E (-10)^20

Posted from my mobile device


The choices have something wrong.

Actually, the choice A and the choice E are same each other.
Without consideration of choices, the least value is -10^20.

However, if we should choose one from the choices, the answer could be D. It is the lease one from possible values.

Happy Studying !!!
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Re: Maths  [#permalink]

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New post 23 Aug 2017, 08:56
kartik0604 wrote:
From the consecutive integers -10 to 10 inclusive 20 integers are randomly chosen with repetition allowed. What is the least value of the product of the 20 integers?
A (-10)^20
B (-10)^10
C 0
D (-10)^19
E (-10)^20

Posted from my mobile device


Hi,

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

E should be \(-10^{20}\)
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Re: Maths  [#permalink]

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New post 23 Aug 2017, 09:25
1
Given number set from -10,-9,-8...0,1,2....9,10

So since we need the least product it has to be negative. so it has to be (-10)^19 * 1 = (-10)^19
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Re: Maths   [#permalink] 23 Aug 2017, 09:25
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