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Set A includes consecutive integers from -10 to 10 (inclusive) [#permalink]
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hathunguyen254 wrote:
GreenlightTestPrep wrote:
KarunMendiratta wrote:
Set A includes consecutive integers from -10 to 10 (inclusive). 20 integers are randomly selected (repetition is allowed) from the set.

Quantity A
Quantity B
Least possible value of the product of all selected integers
0


A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given


Choose nineteen 10's and one -10
So, the product = [(10)^19][-10]
Notice that [(10)^19] is POSITIVE, which means [(10)^19][-10] is NEGATIVE.

So, [(10)^19][-10] = [(10)^19][-1][10]
= -(10)^20
= E

So, we have:
QUANTITY A: -(10)^20
QUANTITY B: 0

Answer: B


I'm a bit confused when it says "consecutive integers". How can the set contains nineteen 10s and one (-10) if they are consecutive integers?


Hi, I think you mis-undertood the question.
The question says, 20 integers are randomly selected (repetition is allowed)

Set A has integers from -10 to 10 only (counted once), but the question wants us to select 20 integers with repetition allowed.
So, we can select -10, -10, -10, ......... (20 times)

As per the question,
The product has to be least. So, we will pick -10 (19 times) and greatest positive integer to make it least i.e. +10
So, Least product = \(-(10)^{19}(10) = -(10)^{20}\)
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Re: Set A includes consecutive integers from -10 to 10 (inclusive) [#permalink]
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Hi, I think you mis-undertood the question.
The question says, 20 integers are randomly selected (repetition is allowed)

Set A has integers from -10 to 10 only (counted once), but the question wants us to select 20 integers with repetition allowed.
So, we can select -10, -10, -10, ......... (20 times)

As per the question,
The product has to be least. So, we will pick -10 (19 times) and greatest positive integer to make it least i.e. +10
So, Least product = −(10)19(10)=−(10)20−(10)19(10)=−(10)20
_________________
I hope this helps!

Regards:


Thanks a lot, I got it !
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