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EventHorizon

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Everything posted by EventHorizon

  1. Inspired by I thought I would give my own. 2,4,3,1,6,5,7 3,7,5,6,1,8,2,4 2,7,6,3,9,4,8,5,1 1) What would it be for 10 numbers? What method am I using? Here's another one (but it is (hopefully) harder, so I'll give all of them up to 9): 1 1,2 1,2,3 4,1,3,2 3,1,4,5,2 6,3,2,5,4,1 1,4,3,2,6,5,7 5,7,8,6,4,1,3,2 4,5,8,6,2,7,3,1,9 2) What is the order for 10 numbers? What is the method?
  2. And another excerpt from the page for "0.999..." So since decimal places are the digits after the decimal point, it seems there is one pair of equal decimal expansion representations for each positive integer and each negative integer. And this is exactly the set of all the numbers that fit the OP. Any other terminating decimal expansion will have the pair of representations equal at one or more decimal places. It is interesting that 0 fails in this regard. The method of decimal representation we use has no way of approaching 0 with repeating 9's due to the fact that adding value i
  3. I found a mistake in the derivation, so my equation is wrong. I accidently omitted something which made it much easier to solve, but incorrect.
  4. (Hint: don't look for the curve, just examine the distance between them and how it changes with time) I'll try (again) to find the equation for the path in a bit. Edit: I thought I'd include the last equation I got, but kept failing to verify. So I don't know it if is correct.
  5. So now we just need to determine if it matters if the frictionless tunnel goes through the center of the earth or not.
  6. This means that you can completely ignore the parts of the earth further than you from the center of the earth, so effectively you feel the gravity of standing on a smaller earth. The effective mass of this smaller earth would be d*(4/3)pi*r2^3, where d is the density of the earth (assumed in the OP as homogeneous) and r2 is your distance from the center of the earth. Notice that this effective mass is equal to the mass of the earth multiplied by r2^3/r^3. Plugging the new mass and radius into the equation to calculate gravitation gives f2 = GmE(r2^3/r^3)/r2^2 = (r2/r)*GmE/r^2 = (r2/r)
  7. So it looks like I had paradoxes on the mind when drafting that post. What I meant (in each of the three mentions of the word paradox) was contradiction. I hate when I read through one of my posts and find misused words. Too bad I'm no longer a VIP or I'd be able to fix them. Edit: Hey, I'm a VIP again, thanks rookie1ja.
  8. Didn't notice psykomakia's post. That's what i get for making dinner while puzzling.
  9. Third, fold again using a fan-like method as shown in the diagram below, alternating forwards and backwards folds. Now all the perforation and only the perforation is along that diagonal... so letta rip.
  10. #include <stdio.h> #include <stdlib.h> #include <iostream> #include <memory.h> #include <vector> using namespace std; #define GRIDSIZE 8 enum dir { dir_front = 0, dir_back = 1, dir_up = 2, dir_down = 3, dir_right = 4, dir_left = 5, num_dirs = 6, dir_error = 7, dir_undo = 8 }; dir facing; dir move; vector<int> hist; int displacement = {0,0,-GRIDSIZE,GRIDSIZE,1,-1,0,0}; int startpos; dir startdir; dir tiltarray[] = {dir_error,dir_error,dir_up,dir_down,dir_right,dir_left, dir_error,dir_error,dir_down,dir_up,d
  11. Yeah, it would be easy. So easy in fact... that here it is Found one! _________________ |1 v=v=v=v=v=v=v| ||X|X X X X X X|| |v 6=< v=v=v >=6| ||X X|X|X X|X|X | |6 6=< 6=< 6 >=6| ||X|X X X|X|X X|| |^=^ >=6=< ^=^=^| | X X|X X X X X | |v=v >=6=< v=v=v| ||X|X X X|X|X X|| |6 6=< 6=< 6 >=6| ||X X|X|X X|X|X | |^ 6=< ^=^=^ >=6| ||X|X X X X X X|| |1 ^=^=^=^=^=^=^| ----------------- Found one! _________________ |1 v=v=v=v=v=v=v| ||X|X X X X X X|| |v 6 6=<=1 >=6 6| ||X|X|X X X|X|X|| |6 ^=^ 6=< > ^=^| ||X X X|X|X
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