   The Magic Cafe Forum Index » » Puzzle me this... » » Power problems (0 Likes) Nir Dahan Inner circle Munich, Germany 1390 Posts  Posted: Jan 14, 2007 10:57 am 0 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+y-1 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 Adventures in ASIC digital design Steve Martin Inner circle 1119 Posts  Posted: Jan 14, 2007 04:25 pm 0 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 2n-1. When n is even, that sum is n^2. Therefore, on a 20x20 board there would be 20^3 cubes. Any man who reads too much and uses his own brain too little falls into lazy habits of thinking. Albert Einstein Nir Dahan Inner circle Munich, Germany 1390 Posts  Posted: Jan 14, 2007 04:35 pm 0 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. Adventures in ASIC digital design Steve Martin Inner circle 1119 Posts  Posted: Jan 14, 2007 04:38 pm 0 1) The answer is 8^3. In general, it is (n-2)^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 Any man who reads too much and uses his own brain too little falls into lazy habits of thinking. Albert Einstein Nir Dahan Inner circle Munich, Germany 1390 Posts  Posted: Jan 15, 2007 01:39 am 0 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 Adventures in ASIC digital design drkptrs1975 Elite user North Eastern PA 452 Posts  Posted: Jan 15, 2007 09:54 am 0 I am not sure what you are asking. The Magic Cafe Forum Index » » Puzzle me this... » » Power problems (0 Likes)
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