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From the consecutive integers -20 to 20, inclusive, 12 even integers

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From the consecutive integers -20 to 20, inclusive, 12 even integers  [#permalink]

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New post 12 Aug 2018, 10:35
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From the consecutive integers -20 to 20, inclusive, 12 even integers are randomly chosen with repetitions not allowed. What is the maximum possible value of the product of those 12 integers?
A) \(0\)
B) \(2^3*10!\)
C) \(2^{13}*45*10!\)
D) \((\frac{10!}{5!})^2\)
E) \(2^{12}\)*\((\frac{10!}{4!})^2\)

Source:- Time4education

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Re: From the consecutive integers -20 to 20, inclusive, 12 even integers  [#permalink]

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New post 12 Aug 2018, 10:41
PKN wrote:
From the consecutive integers -20 to 20, inclusive, 12 even integers are randomly chosen with repetitions not allowed. What is the maximum possible value of the product of those 12 integers?
A) \(0\)
B) \(2^3*10!\)
C) \(2^{13}*45*10!\)
D) \((\frac{10!}{5!})^2\)
E) \(2^{12}\)*\((\frac{10!}{4!})^2\)

Source:- Time4education


this is modified OG question: https://gmatclub.com/forum/from-the-con ... 35518.html
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From the consecutive integers -20 to 20, inclusive, 12 even integers  [#permalink]

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New post 12 Aug 2018, 10:46
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PKN wrote:
From the consecutive integers -20 to 20, inclusive, 12 even integers are randomly chosen with repetitions not allowed. What is the maximum possible value of the product of those 12 integers?
A) \(0\)
B) \(2^3*10!\)
C) \(2^{13}*45*10!\)
D) \((\frac{10!}{5!})^2\)
E) \(2^{12}\)*\((\frac{10!}{4!})^2\)

Source:- Time4education


Between -20 and 20 both inclusive - there are a total of 21 even integers.

Since we need to randomly choose 12 even integers and repetitions are not allowed,
the maximum number is possible when we choose 6 negative and positive numbers
at both ends.

20(2^10),18(2*9),16(2*8),14(2*7),12(2*6),and 10(2*5) when multiplied gives us \(2^{6}\)*\((\frac{10!}{4!})\)

Similarly, the negative numbers -20,-18,-16,-14,-12,-10 will also produce the same product.

Therefore, the maximum product(of 12 random even integers) possible is \(2^{12}\)*\((\frac{10!}{4!})^2\)(Option E)
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Re: From the consecutive integers -20 to 20, inclusive, 12 even integers  [#permalink]

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New post 13 Aug 2018, 06:38

Solution


Given:
    • 12 even integers are randomly selected from [-20, 20], without repetitions

To find:
    • Maximum possible value of the product of these 12 even integers

Approach and Working:
    • For the product to be maximum,
      o The magnitude of the product should be maximum, and
      o The sign should be positive
    • By considering the above two statements, we can say that the 12 even integers must be {-20, -18, -16, -14, -12, -10, 10, 12, 14, 16, 18, 20}
      o The extreme values are considered in order to maximise the product
    • Thus, the product of these 12 integers will be equal to, \((-1)^6 * (10^2 * 12^2 * 14^2 * 16^2 * 18^2 * 20^2)\)
    • Since all the integers are even, we can take 2 common from each integer, from this we get the product as,
      o \(2^{12} (5^2 * 6^2 * 7^2 * 8^2 * 9^2 * 10^2)\)
      o \(2^{12} (5 * 6 * 7 * 8 * 9 * 10)^2\)
      o Multiplying and dividing by 4!2, we get the product as, \(2^{12} * (1 * 2 * 3 * 4 * 5 * 7 * 8 * 9 * 10)^2 /4!^2\)
      o Which is equal to, \(2^{12} * 10!^2 /4!^2\)

Therefore, the maximum possible value of the product is \(2^{12} * (10! /4!)^2\)

Hence, the correct answer is option E
Answer: E

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Re: From the consecutive integers -20 to 20, inclusive, 12 even integers  [#permalink]

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New post 13 Aug 2018, 07:55
total 21 possibilities
-20, -18, -16, ..., 0,..., 16, 18, 20
we have to choose total 12 integers which results in maximum product
so if we choose maximum 6 numbers from both ends of above number series then we will have max product
so chosen numbers are -20, -18, -16, -14, -12, -10, 10, 12, 14, 16, 18, 20
2^12 (-10*-9*-8*-7*-6*-5*5*6*7*8*9*10)
2^12 (5*6*7*8*9*10)^2

So, Answer is E
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Re: From the consecutive integers -20 to 20, inclusive, 12 even integers &nbs [#permalink] 13 Aug 2018, 07:55
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From the consecutive integers -20 to 20, inclusive, 12 even integers

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