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25 May 2022, 01:30
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
26% (02:45) correct
74%
(02:23)
wrong
based on 62
sessions
History
Date
Time
Result
Not Attempted Yet
Each vertex of a cube is to be labeled with an integer 1 through 8, with each integer being used once, in such a way that the sum of the four numbers on the vertices of a face is the same for each face. Arrangements that can be obtained from each other through rotations of the cube are considered to be the same. How many different arrangements are possible?
(A) 1
(B) 3
(C) 6
(D) 12
(E) 24
Are You Up For the Challenge: 700 Level Questions
(A) 1
(B) 3
(C) 6
(D) 12
(E) 24
Are You Up For the Challenge: 700 Level Questions
26 May 2022, 09:43
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Hi Can anyone explain the solution and approach followed for this question?
26 May 2022, 10:52
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Asmita9502
Unless I'm missing a clever shortcut, I don't believe that this is a viable GMAC question. The answer ends up being C (or at least I'm pretty sure it does, but it took me about 8 minutes to get there...yikes!!). The solution required a decent amount of (bad) drawings and (even worse) chicken scratch, and I don't think it's worth either of our time for me to post it and you to read it.
I'd say that if you can at least pick up a few of the following elements, you probably have a decent handle on thinking through questions similar to this one that are of the difficulty level you might get on an actual test.
1) The sum of each side needs to be 18. Why? The sum of all the digits is 36 (1+2+3+4+5+6+7+8) and any two opposite sides need to be equal. So, the top and bottom, for example need to be equal and they need to total to 36. That means each has 18.
2) Each digit is counted three times because each vertex is on three faces of the cube. If we add up all the sides, we have a total of 3*36 = 108.
3) You should be cognizant that we're possibly/probably looking at pairs that sum to 9. Maybe you can't articulate why, but the thought should at least cross your mind.
4) You should at least consider trying to figure out what happens if we "anchor" some value to a specific location and start to think about how to fill in the others. For example, if we put 1 at the top right corner of the front face, 2 cannot go at the top left corner of the front face since that would mean that we would still need 15 for the bottom two vertices of the front face (only options are 7 and 8) and also 15 for the top two vertices of the back face (only options would be 7 and 8 but we already used those and can't have them twice). So, 1 and 2 cannot both be on the same edge.
5) You can similarly reason through why 1 and 3 cannot both be on the same edge.
If we put 1 and 4 on the same edge, the other two values on the adjoining faces need to sum to 13, so we can only try a 5-8 pair and 6-7 pair.
If we put 1 and 5 on the same edge, the other two values on the adjoining faces need to sum to 12, so we can only try a 4-8 pair and a 5-7 pair.
If we put 1 and 6 on the same edge, the other two values on the adjoining faces need to sum to 11, so we can only try a 3-8 pair and a 4-7 pair.
1 and 7 on the same edge has already been covered since 1 shares edges with three other digits and we wouldn't have two others to plug into the other two vertices that are adjacent to 1.
The preceding sentence also applies to 1 and 8 on the same edge.
If you get the first three points, you're probably good. If you get 4, giddy up. And if you get 5, you're a nerd!!!
ThatDudeIsANerd
