You sit on ESOC all day puzzling! You know nothing of hard life!

[momuuu]

But I don't even like riddles.

[JakeyBoyTH]

The answer is 1 kg I think. Turns out you shouldnt leave your watermelon in the sun.

How did you work it out? I don't know how losing 1% of mass should equate to a whole kg. Then again I don't know how a watermelon can evaporate in the very environment that it grows in.

[momuuu]

The answer is 1 kg I think. Turns out you shouldnt leave your watermelon in the sun.

How did you work it out? I don't know how losing 1% of mass should equate to a whole kg. Then again I don't know how a watermelon can evaporate in the very environment that it grows in.

Well, at first its 2 kgs. 99% of that mass is water, so 1% of that mass is something else. In other words, the non-water part of the watermelon weighs 20 grams (thats 1% of 2 kgs). If then 98% of the watermelon is water, that means that this non-water part of the watermelon is now 2% of the total mass. It still weights 20 grams however, since that just remains the same. That means that 20 grams is 2% of the total mass. In other words, the total mass is 20*50 = 1 kg

[Vinyanyérë]

Yeah I did this in class. It's the German easy.

My riddle is this:

On Bagshot Island, there is an airport. The airport is the home-base of an unlimited number of identical airplanes. Each airplane has a fuel capacity to allow it to fly exactly 1/2 way around the world, along a great circle. 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 number 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.
(b) Each airplane must have enough fuel to return to airport.
(c) The time and fuel consumption of refueling can be ignored. (so we can also assume that one airplane can refuel more than one airplanes in air at the same time.)
(d) The amount of fuel airplanes carrying can be zero as long as the other airplane is refueling these airplanes. What is the fewest number of airplanes and number of tanks of fuel needed to accomplish this work? (we only need airplane to go around the world)

[spoiler=spoiler]I think it's 6 planes.

1 plane can fly out, hand off whatever fuel it has, and then fly back safely. It can fly out as far as the point 1/4 (1/4 around the world) when doing this, but the further it flies out the less fuel it can give out. There's also no reason to give more fuel than the plane being refueled can take. It turn outs that if we send 2 planes out from 0, the best place for one plane to refuel the other and fly back is at the point 1/6. One plane uses 1/3 of its fuel to fly out, gives 1/3 of its fuel to the other plane, then the last 1/3 to fly back safely. The other plane uses 1/3 of its fuel flying out, then receives 1/3 back and is at the point 1/6 with a full tank.

To do this same technique at the point 1/3, we need 3 planes. This is because we need a plane with a full tank at 1/6 to go to 1/3, hand off its fuel, and fly back to 1/6, at which point it's empty. So it needs another plane to hand off fuel at 1/6 so that it can fly back, and another plane to give it the full tank at 1/6.

I'm not sure how many planes we'd need to do this at the point 1/2, but I think it ends up being quite a few (I haven't completely worked out the math). But we don't need to do that.

The strategy is:

Send a plane out (1 plane)
Give that plane 1/3 fuel at the point 1/6 (requires 1 plane)
Give that plane 1/3 fuel at the point 1/3 (requires 3 planes)
Plane flies from 1/3 to 5/6 with its full tank
Give that plane 1/3 fuel at the point 5/6 (requires 1 plane)[/spoiler]

Mine: You have a box containing 20 strings. At every step, you select two available ends of the strings at random and tie them together (an end is considered unavailable if it's already tied to another end, and available otherwise). You continue to do this until no more ends remain. On average, how many loops (a set of at least 1 string where the first string is tied to the second, the second to the third, ..., and the last string back to the first) do you make by doing this?

Bonus question: How many loops do you make on average with n strings?

[Gendarme]

The point is that you think that the difference between 99% and 98% is small, but in reality, the difference between 1% and 2% is huge. I suppose it is kind of the same thing as people thinking that a probability of 0,0001% and 0,000000001% are more or less the same, even though they differ immensely, thus losing money on lottery tickets and in casinos.

[momuuu]

It's more of a play on words where this 1% decrease doesn't actually mean what it seems to mean. 1% is still not that significant, its just that its phrased in such way that makes it seem like 1% is a lot.

[JakeyBoyTH]

It's more of a play on words where this 1% decrease doesn't actually mean what it seems to mean. 1% is still not that significant, its just that its phrased in such way that makes it seem like 1% is a lot.

Thanks :)

Btw @Gendarme

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was correct with my riddle. Although his working was a bit of hot air, and needed some lift. He was just plane right.

@Vinyanyérë

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was beginning to take off, but in the end he wasn't on the right (flight) path. His reasoning was dragging him.

Don't get in a flap, ill post the solution soon :)

[Gendarme]

I am pretty sure I was wrong, actually.

[iNcog]

-- deleted post --

[Gendarme]

where does the OP title come from

Gendarme: Taunt 25
Gendarme: Taunt 25
Gendarme: Taunt 25
Gendarme: Taunt 25
Gendarme: Taunt 25
Gendarme: Taunt 25
Gendarme: Taunt 25
Gendarme: Taunt 25
Gendarme: Taunt 25
Gendarme: 2525
Gendarme: Oops.
Gendarme: Taunt 25

[Vinyanyérë]

It's more of a play on words where this 1% decrease doesn't actually mean what it seems to mean. 1% is still not that significant, its just that its phrased in such way that makes it seem like 1% is a lot.

Thanks :)

Btw @Gendarme

notification

was correct with my riddle. Although his working was a bit of hot air, and needed some lift. He was just plane right.

@Vinyanyérë

notification

was beginning to take off, but in the end he wasn't on the right (flight) path. His reasoning was dragging him.

Don't get in a flap, ill post the solution soon :)

Alright I've got it:

[spoiler=spoiler]Label the planes 1, 2, 3. All 3 planes leave at the same time heading clockwise. At the point 1/8, plane 3 gives 1/4 fuel each to planes 1 and 2, then flies back. 1 and 2 now have full tanks. At point 1/4, plane 2 gives 1/4 fuel to plane 1, then flies back. Plane 1 then flies to point 3/4, which it can do since it has a full tank.

While that's going on, planes 2 and 3 refuel and now fly out CCW, timed so that they'll reach 3/4 at the same time as plane 1. At the point 7/8, plane 3 gives 1/4 fuel to plane 2, then flies back. Plane 2 continues on and meets plane 1, at which point is gives over 1/4 fuel, then flies back with plane 1. When plane 3 reaches back, it refuels and flies CCW, meeting plane 1 at point 7/8. Plane 3 gives 1/4 fuel to plane 1, and all 3 fly back together.[/spoiler]

[Gendarme]

Well, technically I was wrong. I was just lucky that I got the number correctly.

Anyways, you can't give 1/4 fuel each at the first refuel, only 1/8. You have 1/4 too much fuel, so you can do a few loops on your way back for fun. That is cool imo.

[Vinyanyérë]

If you travel a 1/8 circle you've used up 1/4 fuel

[Gendarme]

Ah. I thought your fractions were referring to the fuel; not the circumference of the circle.

[JakeyBoyTH]

You are pretty close with that last one.

Here is the *official* solution. Although if you get the correct number then I don't see why it wouldnt be a correct solution. There are always more than one solution to any problem.

[spoiler=spoiler]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. After 1/6 of the circumference, B passes 1/3 of its fuel to C and returns home, where it is refueled and starts immediately again to follow A and C.

C continues to fly alongside A until they are 1/4 of the distance around the world. At this point C completely fills the tank of A which is now able to fly to a point 3/4 of the way around the world. C has now only 1/3 of its full fuel capacity left, not enough to get back to the home base. 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. C reaches A at the point where it has flown 3/4 around the world. 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.[/spoiler]

[Laurence Drake]

[Kappatain]

Yeah that poster is certainly puzzling. I bet the airplane's cousin identifies as a battle helicopter..

[momuuu]

Puzzling indeed but I think it refers to the "I sexually identify as an attack helicopter" meme

[Gendarme]

It became a meme? Lauren OP

[iNcog]

-- deleted post --

[momuuu]

Yeah in the last frame his body is surprisingly feminine.

[JakeyBoyTH]

Yeah in the last frame his body is surprisingly feminine.

Did you just assume it's gender?

[noissance]

@Jerom

notification

Can occam's razor kill schrodinger's cat?

[noissance]

@Jerom

notification

or anyone else
Can't answer the question? ^

[momuuu]

Actually I think it does the opposite. It promotes dualistic approach.