Consider the 1st step, initial number of cubes N3 after removal of 1st set of coloured cubes number of cubes left out is (N - 1)3 hence number of cubes removed in 1st step (i.e with colour 1) is
N3 - (N - 1)3 = 3N2 - 3N + 1
Similarly number of cubes removed in 2nd step (i.e with colour 2) is
Similarly number of cubes removed in 3rd step is (i.e with colour 3) and so on.
= 3(N - 1)2 - 3(N - 1) + 1
Number of cubes remaining after 1st step is (N - 1)3
Number of cubes remaining after 2nd step is (N - 2)3 and so on.
Number of cubes with only face is painted with colour 1 is 3(N - 2)(N - 1) = 3N2 - 9N + 6
Number of cubes with only face is painted with colour 2 is 3 (N - 3)(N - 2) = 3N2 - 15 N + 18
From the given condition (3N2 - 9N + 6) + (3N2 - 15 N + 18) = 6N2 - 24N + 24 = 150 from this we will get N = 7.
Number of cubes left after step 3 is 4 x 4 x 4 = 64
When all the exposed faces are painted with colour 4 then number of cubes with only one face painted is 3 x 2 x 3 = 18
From the observation for 1st step we have seen that number of cubes is 3(N - 2)(N - 1) or in other words 3 times the product of 2 consecutive integer that is satisfied only by 18 which is 3 times of 2 x 3.
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Answer B
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