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All the numbers 2, 3, 4, 5, 6, 7 are assigned to the six faces of a cu

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Joined: 02 Sep 2009
Posts: 58402
All the numbers 2, 3, 4, 5, 6, 7 are assigned to the six faces of a cu  [#permalink]

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09 May 2019, 21:35
15
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Difficulty:

65% (hard)

Question Stats:

57% (02:51) correct 43% (03:13) wrong based on 60 sessions

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All the numbers 2, 3, 4, 5, 6, 7 are assigned to the six faces of a cube, one number to each face. For each of the eight vertices of the cube, a product of three numbers is computed, where the three numbers are the numbers assigned to the three faces that include that vertex. What is the greatest possible value of the sum of these eight products?

(A) 312

(B) 343

(C) 625

(D) 729

(E) 1680

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Joined: 18 Jun 2019
Posts: 6
All the numbers 2, 3, 4, 5, 6, 7 are assigned to the six faces of a cu  [#permalink]

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22 Jul 2019, 20:06
7

Its easy with Equations !!

Let us call the six sides of our cube a,b,c,d,e and f (where a is opposite d, c is opposite e, and b is opposite f).
Thus, for the eight vertices, we have the following products:
abc , abe , bcd , bde , acf , cdf , cef and def.
Let us find the sum of these products:
abc + abe + bcd + bde + acf + cdf + aef + def
Factorizing >
b (ac + ae + cd + de) + f (ac + ae + cd + de)
=(b+f) (ac+ae+cd+de)
=(b+f) (a (c+e) + d (c+e))
=(b+f) (a+d) (c+e) ---- Eqn 1

We have the product. Notice how the factors are sums of opposite faces.

Here is one property -- Product is maximum when numbers are equal.
Ex: consider number 6.
I can write 6 = 5+1 or 4+2 or 3+3
But product of these numbers 5*1,4*2 or 3*3 is maximum only for the case when numbers are equal i.e. 3*3

Similarly,
For 2,3,4,5,6 & 7
We need to make 3 groups (as in equation 1) so that groups are equal (as we learned in above property)
In this case it is (7+2) * (6+3) * (5+4)
=9*9*9
=729 ---

Ans D

Please hit Kudos to encourage !!
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Akshay, New Delhi, India.
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Joined: 18 Aug 2017
Posts: 5014
Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE: Marketing (Energy and Utilities)
Re: All the numbers 2, 3, 4, 5, 6, 7 are assigned to the six faces of a cu  [#permalink]

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09 May 2019, 23:53
Bunuel wrote:
All the numbers 2, 3, 4, 5, 6, 7 are assigned to the six faces of a cube, one number to each face. For each of the eight vertices of the cube, a product of three numbers is computed, where the three numbers are the numbers assigned to the three faces that include that vertex. What is the greatest possible value of the sum of these eight products?

(A) 312

(B) 343

(C) 625

(D) 729

(E) 1680

to maximize value of product ;
(7+2)*(6+3)*(5+4) ; 9*9*9 ; 729
IMO D
Director
Joined: 27 May 2012
Posts: 901
Re: All the numbers 2, 3, 4, 5, 6, 7 are assigned to the six faces of a cu  [#permalink]

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21 Jul 2019, 09:23
Archit3110 wrote:
Bunuel wrote:
All the numbers 2, 3, 4, 5, 6, 7 are assigned to the six faces of a cube, one number to each face. For each of the eight vertices of the cube, a product of three numbers is computed, where the three numbers are the numbers assigned to the three faces that include that vertex. What is the greatest possible value of the sum of these eight products?

(A) 312

(B) 343

(C) 625

(D) 729

(E) 1680

to maximize value of product ;
(7+2)*(6+3)*(5+4) ; 9*9*9 ; 729
IMO D

Hi Archit3110,

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- Stne
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Joined: 18 Aug 2017
Posts: 5014
Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE: Marketing (Energy and Utilities)
Re: All the numbers 2, 3, 4, 5, 6, 7 are assigned to the six faces of a cu  [#permalink]

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21 Jul 2019, 09:43
1
stne ; i did not solve the question considering the product of three nos and then adding the sum of each vertice as then arranging the no on each face of cube would have been a task
the question has asked us to find greatest possible value so to find the largest value need to multiply the greatest no with smallest value and so on ..
total 3 such pairs are there (7+2)*(6+3)*(5+4) ; 729

stne wrote:
Archit3110 wrote:
Bunuel wrote:
All the numbers 2, 3, 4, 5, 6, 7 are assigned to the six faces of a cube, one number to each face. For each of the eight vertices of the cube, a product of three numbers is computed, where the three numbers are the numbers assigned to the three faces that include that vertex. What is the greatest possible value of the sum of these eight products?

(A) 312

(B) 343

(C) 625

(D) 729

(E) 1680

to maximize value of product ;
(7+2)*(6+3)*(5+4) ; 9*9*9 ; 729
IMO D

Hi Archit3110,

Senior Manager
Joined: 25 Jul 2018
Posts: 266
Re: All the numbers 2, 3, 4, 5, 6, 7 are assigned to the six faces of a cu  [#permalink]

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21 Jul 2019, 10:45
3
Posted from my mobile device
Attachments

File comment: PLEASE FIND THE ATTACHED FILE.

I think, logic is that we need to put big numbers next to each other.

(P.S: I got the wrong solution first)

06064B46-6DFD-41AF-9651-B88FD93FD716.jpeg [ 761.3 KiB | Viewed 674 times ]

Re: All the numbers 2, 3, 4, 5, 6, 7 are assigned to the six faces of a cu   [#permalink] 21 Jul 2019, 10:45
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