GK 2016 Q25

Note first that a small cube in the centre of a face of the large cube will only appear on one face while a cube appearing on the edge of a face of the large cube will appear on two faces and a cube appearing at a corner of the face of the large cube will appear on three faces. Hence, the total number of white faces on the edge of the large cube is an even number and the total number of white faces on the corners of the large cube is a multiple of 3. The five faces shown contain 1 centre white face from 1 small white cube, 12 edge white faces and 5 corner white faces. Therefore, since the total number of white faces on the corners is a multiple of 3, the missing face contains 1 or 4 white faces at its corners. None of the options contains 4 white corners so the missing face contains one white corner as in options A and E, making 6 in total. These 6 faces come from 6 ÷ 3 = 2 small white cubes. Both options A and E have two white faces on their edges, making 14 in total over the six faces from 14 ÷ 2 = 7 white cubes. Hence the number of white cubes whose positions we know is 1 + 7 + 2 = 10. The large cube is made with 12 small white cubes so there are still two more to be placed. One can be at the centre of the large cube and the only place the remaining cube can be is at the centre of the missing face. Therefore, the missing face contains one centre white face, two edge white faces and one corner white face. Hence, the missing face is A. (This proof shows that the only possible missing face for such a cube is face A. It is left to the reader to check that the five given faces and face A can indeed be fitted together consistently to form the faces of a cube.)