TimeSpaceLightForce

Thats it! journey complete..thanks

Five balloonist discharges the same amount of hellium to decent back to their camp below 1000 ft. below for breakfast. One of them left his sandbag while the others still got 1 ,2 ,3 ,4 Lbs of sandbag so that the respective speeds is 1 ,2 , 3, 4, 5. fps In order to arrive altogether at the same time to the campsite they may use the ropes to pass or exchange their sandbags with the others at any time.Starting all on the edge of 1000 ft plateau, how would they plan the return with minimum passes or exchanges? (()) (()) (()) (()) (()) -\/__\/__\/__\/__\/

All right!

Five balloonist will travel up to a 1000 ft. high plateau to see the sunrise. Each carry a rope and a 1, 2, 3, 4, 5 Lbs sandbag so that they ascend at 5, 4, 3, 2, 1 ft/sec respectively.(the hellium balloons are similar) In order to arrive altogether at the same time to the top they may use the ropes to pass or exchange their sandbags with the others at any time. Starting all on the ground, how would they plan the trip with minimum passes or exchanges? (()) (()) (()) (()) (()) \/__\/__\/__\/__\/

Nice puzzle .. Can we label the rooms, boxes, real number - 00 to 99 ? So that room 50 has a box with "50" written outside and "50" written inside..

DUH DUH DUH d u h Duh Duh Duh d u h . . . . -LvB

i was thinking that triangle intersection is overlapping of area.. line intersection is overlapping of point

flip the card..

he can do much better than ten

No, the boy only lose a coin for making 7 and only the boy places all his coins

The father told his son to make as many non-intersecting triangles with his eight coins. And for every such triangle (3 centers) counted on the table he will pay 1 coin but take all coins used to make those triangles . How many coins shall the boy win or lose?

[spoiler=It has to be different in the following way]In your sketch, the top row circles are 1-1 with the bottom row. It has to be the case that two extra top row circles appear. That won't happen unless at some point (at least in two cases) a top row circle is directly above a bottom row circle. So the top row circles must move up and down in groups. Start with a group of three top row circles nestled down in triangular packing position. Then a group of three top row circles packed against the top of the box. Then three more down, then three more up. Do that 66+ times, to the end of the box. It seems like that permits the top row to be compacted in length. That may be enough to fit two extra (199 = 2 = 201) top row balls.