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

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Director
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Joined: 01 Oct 2017
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WE: Supply Chain Management (Energy and Utilities)
From the consecutive integers -20 to 20, inclusive, 12 even integers  [#permalink]

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12 Aug 2018, 11:35
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Difficulty:

65% (hard)

Question Stats:

59% (02:39) correct 41% (02:43) wrong based on 61 sessions

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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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PKN

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Joined: 02 Sep 2009
Posts: 59561
Re: From the consecutive integers -20 to 20, inclusive, 12 even integers  [#permalink]

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12 Aug 2018, 11:41
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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Joined: 26 Feb 2016
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Location: India
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From the consecutive integers -20 to 20, inclusive, 12 even integers  [#permalink]

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12 Aug 2018, 11:46
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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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13 Aug 2018, 07:38
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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

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Joined: 20 Jul 2018
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Re: From the consecutive integers -20 to 20, inclusive, 12 even integers  [#permalink]

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13 Aug 2018, 08:55
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

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Hasnain Afzal

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Re: From the consecutive integers -20 to 20, inclusive, 12 even integers   [#permalink] 13 Aug 2018, 08:55
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