Consider three identical airplanes starting at the same airport. Each plane has a fuel tank that holds just enough fuel to allow the plane to travel half the distance around the world. These airplanes possess the special ability to transfer fuel between their tanks in mid-flight. What are the maximum around the world trips that airplane1 can make?

I found this question on interviewbit and this really blows me off. I frankly have no idea how to proceed. I thought answer would be 0 since it apparently seems impossible but answer is 1.

EDIT: This is not the same as the original question, but it is properly sourced. OP can decide what to keep.

A group of aeroplanes is based on a small island. The tank of each plane holds just enough fuel to take it halfway around the world. Any desired amount of fuel can be transferred from the tank of one plane to the tank of another while the planes are in flight.

The only source of fuel is on the island, and for the purposes of the problem it is assumed that there is no time lost in refuelling either in the air or on the ground.

What is the smallest number of planes that will ensure the flight of one plane around the world on a great circle, assuming the planes have the same constant ground speed and rate of fuel consumption and that all planes return safely to their island base?

Source: http://xn--webducation-dbb.com/wp-content/uploads/2019/01/Martin-Gardner-More-mathematical-puzzles-and-diversions-Pelican-books-1961.pdf [Original text page number 39]

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    $\begingroup$ @dan04 Doesn't matter. They have half enough fuel no matter how you define it. $\endgroup$ Commented Aug 9, 2022 at 22:06
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    $\begingroup$ If you are interested in mathematical puzzles, the original version of this puzzle was published in one of Martin Gardners books (don't remember which one, most likely one of "The Xth Scientific American Book of Puzzles and Games"). $\endgroup$ Commented Aug 10, 2022 at 8:40
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    $\begingroup$ Per the InterviewBit TOS, "You are prohibited from... copying or duplicating in any manner any of the Scaler Content or other information available from the Platform", thus I'm voting to close this under our policy against sharing puzzles without the creators' consent $\endgroup$
    – bobble
    Commented Aug 11, 2022 at 4:28
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    $\begingroup$ @WGroleau, It can't be copyright violation: this is an old trivial task, I've seen its variations elsewhere long ago. $\endgroup$
    – Zeus
    Commented Aug 11, 2022 at 7:43
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    $\begingroup$ @Charlie content can be provided for free but with terms that it can't be shared. Elsewhere the Terms define "Scaler" to mean "InterviewBit" and "Content" to be things including any text on the site. The copyright is not the general idea of the problem but rather the specific formulation/word choice/presentation. $\endgroup$
    – bobble
    Commented Aug 11, 2022 at 13:50

6 Answers 6


Plane 1 can make 1 trip around the world by following the below process:

Plane 1 and Plane 2 take off in the same direction at the same time. Once they reach 1/4 of the way around the globe, plane 2 transfers all remaining fuel to plane 1. At this point Plane 1 has a full tank of fuel (half globe), Plane 2 flames out, and Plane 3 is still at the airport.

Once that step is met, then:

Plane 1 continues to fly until going half way around the world. With the extra half tank of fuel from plane 2, Plane 1 has a half tank of fuel left. Plane 3 departs the airport at this time, flying in the opposite direction.

Step 3:

Plane 3 and Plane 1 meet up a quarter of the way further around the globe (so plane 1 is 75% of the way around and flying on fumes). At this point Plane 3 transfers all remaining fuel (half a tank, enough for a quarter of the way around the world) to Plane 1.

At this point:

Plane 1 can use the transferred fuel from Plane 3 to complete it's circumnavigation. Plane 2 is flamed out (glided to safety or crashed) to one direction of the airport, Plane 3 is flamed out 25% in the opposite direction.

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    $\begingroup$ Now the trick is to find 3 spots on the surface of the earth that are each 90 degrees apart from the next on a great circle - and all on land, preferably near an airport. (Hard enough to find a pair of points on land 180 degrees apart, as most land has water at its antipodes. To then find a starting point 90 degrees from both would be another challenge...) $\endgroup$ Commented Aug 10, 2022 at 13:49
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    $\begingroup$ @DarrelHoffman The planes transfer fuel mid-air, and there's no requirement to actually land any of the planes, rather than simply ejecting when the fuel tank is empty. You don't need any land at all except the starting airport. An interesting geography question, though. $\endgroup$ Commented Aug 10, 2022 at 15:52
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    $\begingroup$ Of course this only works if you make the unrealistic assumption that fuel burn is independent of current fuel load. $\endgroup$ Commented Aug 10, 2022 at 17:13
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    $\begingroup$ @PeterGreen Yeah, definitely don't suggest asking this question in Aviation.SE. Not only because of stuff like that, but also I think they'd be somewhat horrified that we're just ejecting and crashing 2 of the planes to make this happen... $\endgroup$ Commented Aug 10, 2022 at 17:19
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    $\begingroup$ @DarrelHoffman Meanwhile, the RAF wants you to hold their beer. see Operation Black Buck. $\endgroup$
    – Aron
    Commented Aug 11, 2022 at 3:15

My first attempt resulted in crashes. Thanks Ross for pointing it out.

Crashes are bad. We must land all planes.

Problems like this require you thinking in the right units. Here's a revised solution:

Assumptions: A tank holds 180 units of fuel.
Each degree of travel consumes 1 unit of fuel.
Planes always travel 1 degree a minute,
Fuel transfer is instantaneous.
Denote a plane as A(position in degrees, Fuel)

1: A( 0, 180), B( 0, 180), C( 0, 180). (t = 0 m)
    A, B, C all take off at the same time, and
2: A( 45, 135), B( 45, 135), C( 45, 135). (t = 45 m)
    fly 45 degrees.
3: A( 45, 180), B( 45, 180), C( 45, 45). (t = 45 m)
    C transfers 45 units of fuel to each of A and B.
4: C( 0, 180). (t = 90 m)
    C returns and refuels
5: A( 90, 135), B( 90, 135) (t = 90 m)
    A and B carry on to 90.
6: A( 90, 180), B( 90, 90) (t = 90 m)
    B transfers 45 fuel to A
7: B( 0, 180) (t = 180 m)
    B returns and refuels
8: A(270, 0) (t = 270 m)
    A carries on to 270.
9: A(270, 0) C(270, 90) (t = 270 m)
    C takes off in the other direction at t = 180 to meet A at 270.
10: A(270, 45) C(270, 45) (t = 270 m)
    C transfers 45 units of fuel to A.
11: A(315, 0) C(315, 0) (t = 315 m)
    C and A carry on to 315.
12: A(315, 0) B(315, 135) C(315, 0) (t = 315 m)
    B takes off at t=270, and travels to 315.
13: A(315, 45) B(315, 45) C(315, 45) (t = 315 m)
    B transfers 45 units to each of A and C,
14: A(360, 0) B(360, 0) C(360, 0) (t = 360 m)
    They all have just enough fuel to get home.
The main trick is that the question specifies tank size, not fuel amount. So refueling is allowed.

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    $\begingroup$ This is what I imagine when reading the question. Nice one saving all planes! $\endgroup$
    – justhalf
    Commented Aug 10, 2022 at 2:57
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    $\begingroup$ Knowing the intended answer is 1 I think this is not the intended solution. Because this solution has actually as answer infinite because the procedure can be repeated as much as you like. So refuelling is probably not allowed $\endgroup$
    – Ivo
    Commented Aug 10, 2022 at 6:27
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    $\begingroup$ You could probably do it with two planes if plane B can land at other airports along the way and refuel while plane A circles overhead. I think the question needs rewording. $\endgroup$ Commented Aug 10, 2022 at 7:44
  • $\begingroup$ The "you can refuel plane 3" is a good insight. $\endgroup$ Commented Aug 10, 2022 at 11:06
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    $\begingroup$ @Ivo Yes, It is infinitely repeatable. If you can do it once, and land all planes, then we can simply do it again, as many times as we want. However, I actually think the question is ambiguous. I went with an implicit assumption that we need to land all planes, and without that restriction being explicit, it's all open for debate. $\endgroup$ Commented Aug 10, 2022 at 13:57

For funsies, let's see how far they could go using asparagus staging.

That is, plane 3 is continuously topping up plane 2, who is continuously topping up plane 1.

This means that plane 3 is burning fuel 3x as fast while the other two are topped up, so if we say that C is the circumference of the earth, plane 2 and plane 1 make it C/6 before losing plane 3

now plane 1 and plane 2 have a full tank, and plane 2 is topping up plane 1 continuously. That means we can go C/4 distance before plane 2 runs out, for a total of 5C/12. Almost halfway! Unfortunately, with a full tank, plane 1 can only travel up to C/2 further, so will be 11C/12, just one twelfth away from a full navigation.

In fact, you would need 4 planes to complete the circumnavigation. With this method, you have the series $$ \frac{1}{2} \sum_{i=1}^{n}\frac{1}{a}$$ for n planes. The fourth term here gives 25/24, which is just over 1 circumnavigation! Whoo!

And as n goes to infinity, because the series diverges you can make infinitely many circumnavigations!

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    $\begingroup$ I love it. KSP was my first thought when I read the question. $\endgroup$ Commented Aug 11, 2022 at 21:06
  • $\begingroup$ I think this is how the Vulcan bomber made it to the Falklands. $\endgroup$
    – Tony Ennis
    Commented Aug 12, 2022 at 18:20

I think my solution uses the least total fuel and pilot/aircraft time, plus the task is completed and

everybody gets home safely, meaning its maximum repeatability count is primarily limited by fuel supply, but still higher than any alternative answers.

First, we need to clarify the question:

What, exactly, does flying around the world mean?
Does it mean that you have to pass through all 360 degrees of longitude?

Assuming that suggested clarification is accepted, here's the solution:

Plane 1 flies to the nearest pole, does a tight circle (thus, flying around the world), and returns.

Why does this work?

Assume for a moment the plane is on the equator, as far away from the poles as possible.
Because the Earth bulges out a bit at the equator, having enough fuel to go halfway around the world there is more than enough to get to the pole and back, enough for that circle at the pole.
If the airport is not on the equator, even less fuel is required to reach the nearest pole.

If the suggested clarification is accepted, but the only fuel available is what's in full aircraft tanks,

then the maximum number of times this can be repeated is 3.

If the suggested clarification is not accepted:

If the challenger adds that you have to travel through all degrees of latitude too, ask if the airport is on the equator. If so, follow the above procedure, refuel, and do the same with the other pole. If the challenger clarifies that the challenge requires completing a "great circle," see other answers.

  • $\begingroup$ The clarification doesn't actually change anything at all, since the amount of fuel is 1/2 of the distance around the world. If you assume you can fly around a pole, then the plane has 1/2 of the polar circle in fuel. If it's a great circle around the equator, the plane has 1/2 that amount of fuel. The amount of fuel is a percentage of the distance needed to be flown, no matter what that distance is $\endgroup$ Commented Aug 12, 2022 at 0:26
  • $\begingroup$ @AnthonyIngram-Westover If you're on the equator with 1/2 the distance around the equator in fuel, you can make it up around a pole and back. $\endgroup$
    – WBT
    Commented Aug 12, 2022 at 0:29
  • $\begingroup$ Right, but the goal is to travel x miles where x is the distance around the world (however you choose to define that) and there is enough fuel to travel x/2 miles. However you define x (distance to travel around the world) the amount of fuel is half that. $\endgroup$ Commented Aug 12, 2022 at 16:43
  • $\begingroup$ @AnthonyIngram-Westover Where are you reading that in the question as originally stated? You appear to be assuming a constraint which isn't actually there: what you assume to be a constant x is actually x and y and the assumption x=y. Clarification questions allow the puzzle-presenter the opportunity to add constraints; they may or may not. Challenging those assumptions is key to the kind of out-of-the-box thinking this answer represents. $\endgroup$
    – WBT
    Commented Aug 12, 2022 at 20:39
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    $\begingroup$ By your definition, a plane going from North pole to south pole and then continuing to north pole did not went around the world. This feels strange... For me a more formal definition of "around" is that if you project the object you go around (here the earth) on the 2D space of your loop (the path taken by your plane), then all points of the projection are inside the loop. With this definition, your solution doesn't work sadly. $\endgroup$ Commented Aug 12, 2022 at 21:19

I'm late to the party, but I think the accepted answer is wrong

all planes fly west
at 1/8 of the trip: plane 3 fills the other planes, and just make it back with the remaining 1/4 tank - it refuels at the airport and flies west again
at 1/4 of the trip: plane 2 fills plane 1 and returns; When plane 3 and plane 2 meet, they redistribute their fuel and fly back At exactly half the trip they refuel at the airport and fly east
at 5/8 of the trip: plane 3 fills plane 2, heads back, and refuels
at 3/4 of the trip: plane 2 shares fuel with plane 1, and both fly west
at the same time plane 3 flies east again, and at 7/8 is in time to share enough fuel with both other planes.
al planes land safely and can repeat the process indefinitely (as long as there is fuel at the airport.
So the answer is, an infinite number of times. (as long as there is fuel at the airport. )


General Solution Where All Airplanes Survive

If we don't want to destroy any planes, this problem leads to a nifty insight - not surprising since it was shared by Martin Gardner. I will answer his variant asking for the number of planes needed so the treatment is more general.

The general template for the strategy is similar to Chris Cudmore's answer ( https://puzzling.stackexchange.com/a/117441/81255 ) : there is some send-off help where the leading airplane receives fuel, followed by a solo leg for the leader, finished by a last-stretch help where the leading airplane receives some fuel again.

A Note on Fuel-Transfer

Say there are $n$ planes flying: a chosen leader, $L$, to make the full journey around the world, and $(n-1)$ helpers. Since fuel-transfer takes $0$ time between any 2 airplanes, one can make infinitesimal transfers between any two airplanes in quick succession. Quick succession with $0$ time in between is effectively simultaneous fuel-transfer. So we can assume that many airplanes can send/receive fuel simultaneously and continuously from any number of airplanes.

Units of Measurement

Let's measure distance in fraction of great-circle length, and fuel in fraction of tank-capacity. Since time and velocity aren't specified in the problem, we can simply measure time as distance. So we can say "a plane loses fuel at a rate of $2x$" to mean "if the plane travels a distance $x$ (fraction of great-circle), its fuel reduces by $2x$ (fraction of tank-capacity)"

Burn Factor

When an airplane is helping, it loses fuel at a rate of $2bx$ for some $b>1$, since it is spending fuel for itself and at least one other airplane. Let's call $b$ the burn factor. For example, if a single helper is helping $(n-1)$ other airplanes, then $n$ airplanes are flying on effectively 1 tank and the burn factor, $b$, is $n$. The burn factor here is essentially the number of planes the helper's tank is fueling.


  1. On the send-off stretch, the first helper plane fuels all other planes for a distance $r_1$ and then returns to refuel, while the the second helper plane takes over and fuels the remaining for a distance $r_2$ and returns, and so on, with airplane $i$ fueling the remaining pack for a distance $r_i$ before returning to refuel, till only the leader, $L$, is left.

  2. $L$ flies solo for a distance.

  3. The refueled helpers fly in the opposite direction, rewinding the send-off pattern: the first refueled helper rewinds the tape of the last returned helper. Clearly they can help $L$ the same distance on the last-stretch as in the send-off stretch and all planes land.


We just need to consider the total send-off helping distance since the last-stretch helping distance is the same.

The first helper will lose fuel at the rate of $2b_1 x$ when helping, with a burn factor $b_1 = n$. Helper plane $i$ will lose fuel at the rate of $2 b_i x$ when it is helping, with $b_i = b_{i-1} - 1$ since it is fueling one less plane than the previous helper. Every helper plane loses fuel at the rate of $2x$ when returning to refuel.

Helper plane $i$ will have a full tank when it starts helping, when helper $(i-1)$ turns back to refuel. It has one less plane to help compared to $(i-1)$, but it will have to travel a longer distance back to the island to refuel.

If every helper helps maximally, it will have just enough fuel to reach the island when it turns back. The first helper will have $1 - 2 b_1 r_1$ fuel left in the tank when it turns, which should be exactly $2r_1$ to make it back.The second helper will need to have $2(r_1 + r_2)$ fuel to make it back, and so on:

$$ \begin{array}{ll} 1 - 2 b_1 r_1 = 2 r_1, \\ 1 - 2b_2 r_2 = 2 (r_1 + r_2) = (1 - 2b_1 r_1) + 2r_2, \\ \vdots \\ 1 - 2b_{n-1} r_{n-1} = 2(r_1 + r_2 + \cdots + r_{n-1}) = (1 - 2b_{n-2}r_{n-2}) + 2r_{n-1}, \end{array} $$ or, more succinctly, $$ 1 - 2b_i r_i = 1 - 2b_{i-1} r_{i-1} + 2 r_i, \quad 2 \leq i \leq n-1. $$ Since $b_i = b_{i-1} - 1$ and $b_1 = n$, it is straightforward to see that

$$ r_i = \frac{1}{2(b_1 + 1)} = \frac{1}{2(n+1)} \forall i < n. $$ So every helper plane pushes itself and the remaining pack the same distance forward. Since there are $n-1$ helpers, the total send-off helping distance is $$ \sum_{i=1} ^{n-1} r_i = \frac{n-1}{2(n+1)}. $$

The last-stretch helping distance is identical since it is just a tape-rewind of the send-off, resulting in a total helping distance of $$ \frac{n-1}{n+1}. $$ Clearly this has to be at least $1/2$ for $L$ to be able to fly the rest by itself. So $$ \frac{n-1}{n+1} \geq \frac{1}{2} \implies n \geq 3. $$

Addendum 1

Interestingly, when $n=3$, all planes will be empty upon reaching the finish-line: on the last stretch, $L$ will be met just in time by one helper exactly when it is empty, and the first helper will be met by the second exactly when it is empty. So the total fuel used in this case will be 5 tanks.

Addendum 2

A seemingly viable alternative, where all helpers simultaneously help $L$ (but not each other), is actually never viable for any $n$, since the ratio of how much a helper's fuel goes into flying itself versus flying another plane is worse (larger).

Hope this amuses/helps!


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