(Close Window) 
Topic: Power problems 


1) you got a 10x10x10 cube made up of 1x1x1 white cubes. Someone comes and paints the outer side of the big cube with black paint. All the cubes with a black side fall to the ground. How many are left? 2) you have a 20x20 chess like board. Imagine the x and y axis as its sides. on 1,1 there is 1 cube. on 1,2 and 2,1 there are (on each) stacked 2 cubes on 1,3 2,2 and 3,1 there are (on each ) stacked 3 cubes in general: on coordinate x,y you got stacked x+y1 cubes How many cubes on total are on the 20x20 board?  Try finding an elegant solution for both problems, not just counting. you will need all your "powers" for this... Nir 


2) For a square with side n, the total number of cubes is n x (sum of cubes on the x=y diagonal). The sum of the cubes on the x=y diagonal is the sum of the odd numbers from 1 to 2n1. When n is even, that sum is n^2. Therefore, on a 20x20 board there would be 20^3 cubes. 


Steve  the answer is correct, but could you find a way without calculating the sum of the diagonal? The answer I am looking for is technically the same, but just a different prespective on the problem. 


1) The answer is 8^3. In general, it is (n2)^3 (for any n). 2) The answer is 20^3. In general it is n^3 (providing n is even). I arrived at my answer to 2) by noting that the total number of cubes is 20 x (sum from 1 to 20) + 20 x (sum from 0 to 19), which is equivalent to 20 x (sum of odd numbers from 1 to 39). The sum of odd numbers from 1 to 39 is in fact 20^2. Hence total = 20^3 


I guess I am being a bit petty. What I meant was that in question 2  if you "fold" everything above the 20 cube height over the long diagonal  you will get a perfect 20x20x20x cube. For answer 1  you are perfectly correct. You will be surprised how many people try to count the cubes falling on the floor, especially if you ask how many there are on the floor and not how many are left... (maybe I should have asked that instead...) nir 


I am not sure what you are asking. 