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# Maths

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Intern
Joined: 18 Aug 2017
Posts: 2

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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

Posted from my mobile device
Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 7224
GMAT 1: 760 Q51 V42
GPA: 3.82

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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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Math Expert
Joined: 02 Aug 2009
Posts: 7556

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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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Manager
Joined: 22 May 2015
Posts: 108

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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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Consistency is the Key
Re: Maths   [#permalink] 23 Aug 2017, 09:25
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# Maths

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