The airport is the home-base of an unlimited of identical airplanes. The planes have the ability to refuel in flight without loss of speed or spillage of fuel.
Though the fuel is unlimited, the island is the only source of fuel. What is the fewest of aircraft necessary to get one plane all the way around the world assuming that all of the aircraft must return safely to the airport? How did you get to your answer?
Notes: a Each airplane must depart and return to the same airport, and that is the only airport they can land and refuel on ground. What is the fewest of airplanes and of tanks of fuel needed to accomplish this work?
The world's hardest riddle
As per the puzzle given above The fewest of aircraft is 3! Imagine 3 aircraft A, B and C. A is going to fly round the world. All three aircraft start at the same time in the same direction. But the first 'auxiliary' aircraft reaches it in time in order to refuel it, and both 'auxiliary' aircraft are the able to return safely to the home base.
Now in the same manner as before both B and C fully refueled fly towards A. Again B refuels C and returns home to be refueled. All 3 aircraft can safely return to the home base, if the refueled process is applied analogously as for the first phase of the flight.
Here’s your chance to solve the world’s hardest riddle
Both eggs are identical. So if the first egg breaks maximum we can have x-1 drops and so we must always put the first egg from height x. So we have determined that for a given x we must drop the first ball from x height.
Taking an example, lets say 16 is my answer. That I need 16 drops to find out the answer. Lets see whether we can find out the height in 16 drops. First we drop from height 16,and if it breaks we try all floors from 1 to The reason being if it breaks at 32nd floor we can try all the floors from 17 to 31 in 14 drops total of 16 drops.
9 of the toughest riddles ever
Now if it did not break then we have left 13 drops. From the above table we can see that the optimal one will be needing 0 linear trials in the last step. So the answer is: 14 Drop first orb from floors 14, 27, 39, 50, 60, 69, 77, 84, 90, 95, 99, Difficulty Popularity You are the ruler of a medieval empire and you are about to have a celebration tomorrow. The celebration is the most important party you have ever hosted. You've got bottles of wine you were planning to open for the celebration, but you find out that one of them is poisoned.
The poison exhibits no symptoms until death. Death occurs within ten to twenty hours after consuming even the minutest amount of poison.
You have over a thousand slaves at your disposal and just under 24 hours to determine which single bottle is poisoned. You have a handful of prisoners about to be executed, and it would mar your celebration to have anyone else killed.
What is the smallest of prisoners you must have to drink from the bottles to be absolutely sure to find the poisoned bottle within 24 hours? Bonus points if you worked out a way to ensure than no more than 8 prisoners die. all bottles using binary digits. As each prisoner to one of the binary flags.
Hardest riddle ever
Prisoners must take a sip from each bottle where their binary flag is set. Here is how you would find one poisoned bottle out of eight total bottles of wine. If none die, bottle 1 is bad.
With ten people there are unique combination so you could test up to bottles of wine. Each of the ten prisoners will take a small sip from about bottles. Each sip should take no longer than 30 seconds and should be a very small amount. Small sips not only leave more wine for guests. Small sips also avoid death by alcohol poisoning.
As long as each prisoner is administered about a milliliter from each bottle, they will only consume the equivalent of about one bottle of wine each. Each prisoner will have at least a fifty percent chance of living.
There is only one binary combination where all prisoners must sip from the wine. If there are ten prisoners then there are ten more combination where all but one prisoner must sip from the wine. By avoiding these two types of combination you can ensure no more than 8 prisoners die. One viewer felt that this solution was in flagrant contempt of restaurant etiquette. The emperor paid for this wine, so there should be no need to prove to the guests that wine is the same as the label. I am not even sure if ancient wine even came with labels affixed.
However, it is true that after leaving the wine open for a day, that this medieval wine will taste more like vinegar than it ever did. C'est la vie. Difficulty Popularity Outside a room there are three light switches.
One of switch is connected to a light bulb inside the room. You are allowed to set each switch the way you want it and then enter the room note: you can enter the room only once Your task is to then determine which switch controls the bulb?? Set the first switches on for abt 10min, and then switch on the second switch and then enter the room. Three cases are possible 1.
Difficulty Popularity Four friends need to cross a dangerous bridge at night. Unfortunately, they have only one torch and the bridge is too dangerous to cross without one.
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The bridge is only strong enough to support two people at a time. Not all people take the same time to cross the bridge. Times for each person: 1 min, 2 mins, 7 mins and 10 mins. What is the shortest time needed for all four of them to cross the bridge? How long would that take? Is that it? That would make this question too simple even as a warm up question. To reduce the amount of time, we should find a way for 10 and 7 to go together. If they cross together, then we need one of them to come back to get the others.
That would not be ideal. How do we get around that? Maybe we can have 1 waiting on the other side to bring the torch back.
World's hardest riddles
Ahaa, we are getting closer. The fastest way to get 1 across and be back is to use 2 to usher 1 across. Difficulty Popularity Hussey has been caught stealing goats, and is brought into court for justice. The judge is his ex-wife Amy Hussey, who wants to show him some sympathy, but the law clearly calls for two shots to be taken at Hussey from close range. To make things a little better for Hussey, Amy Hussey tells him she will place two bullets into a six-chambered revolver in successive order.
She will spin the chamber, close it, and take one shot. If Hussey is still alive, she will then either take another shot, or spin the chamber again before shooting.
Hussey is a bit incredulous that his own ex-wife would carry out the punishment, and a bit sad that she was always such a rule follower. He steels himself as Amy Hussey lo the chambers, spins the revolver, and pulls the trigger.