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Let there be $n$ cars on a track and non are driving. If we sum up all the fuel in all cars on the track then one car can travel one round around the track. Prove that you can pick always a car that can drive one whole round around the track by taking the fuel of cars that he passed (only he is moving, others aren't).

e.g: let the $i-th$ car have $a_i$ fuel. Now let $a_1+a_2+...+a_n=x$. $x$ is the amount of fuel put in one car so that it can travel exact one round!

My part: Let all cars face in one direction. It's easy to get that there always must exist one car that has enough fuel to get to the car infornt of him (otherwise the sum of all fuel wouldn't be enough to travel the whole track). After I got that I thoguht that maybe induction could help, bcs after he gets to the car infornt of him we can ignore that car and look at $n-1$ cars. But I am not sure if that is going to help bcs that the chosen car needs to travel the remaining distance because we can't switsch cars (no matter how little he traveled to get to the car infornt of him, if we pick another car it needs to travel the whole distance which is impossible because we used some fuel to get to the car infornt of him).

Sorry if my english is bad and thanks for your help.

CryoDrakon
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1 Answers1

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Add a new car to the track that has enough gas in its tank to make a full round on its own. Let this car (virtually) make a round and during that also refuel from the cars it passes. Then after one round this car has exactly the same amount of fuel as initially. During its round, there was some point $x$ where its fuel was minimal, say $y$. This must happen at one of the given cars (and immediately before it takes that car's fuel). If the car originally at $x$ tries to make a round instead, it will never run dry - this is because it will always have a constant amount less than the "virtual" car would have, namely so that the minimum is taken exactly upon completing the round, hence precisely when the last drop is spent.

Hagen von Eitzen
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  • what was ur motivation , have u encountered this type of question before – avz2611 Feb 03 '15 at 17:36
  • @avz2611: Are you asking why Hagen wrote this answer? – hmakholm left over Monica Feb 03 '15 at 17:36
  • no im asking what was his thought process – avz2611 Feb 03 '15 at 17:37
  • @avz2611: I would say it's a pretty straightforward problem $-$ no prior experience necessary. My solution was similar, but instead of Hagen's supercar, I decide to pick a car at random and imagine it going all the way around, allowing its fuel to go negative if necessary. Then pick the car at which my car's fuel was at a minimum. – TonyK Feb 03 '15 at 17:40
  • I edited the question a bit, look at it now. If there was a car that could make a whole round by it self then it's trivial, but what if it has less than $x$ of fuel? – CryoDrakon Feb 03 '15 at 17:46
  • Just a nit: there might be more than one minimum, so "the minimum is taken exactly upon completing the round" is a bit misleading, that is, suggesting that the minimum won't happen anywhere else. – dtldarek Feb 03 '15 at 17:55