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09 May 2019, 21:35
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D
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Difficulty:
75%
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Question Stats:
62% (02:50) correct
38%
(03:04)
wrong
based on 144
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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
(A) 312
(B) 343
(C) 625
(D) 729
(E) 1680
22 Jul 2019, 20:06
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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
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21 Jul 2019, 10:45
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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 6570 times ]
