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This question already has an answer here:

You, 4 other people, and a dog are in a room. You have a bunch of tennis balls in the center of the room. When no one is looking except for the dog, you give one tennis ball to the dog and then you divide the remaining tennis balls into 5 equal piles. You take one of the piles and hide it somewhere else in the room. Each of the 4 other people in the room do this to the remaining stash of tennis balls in the center (1 for the dog, 1/5 of the remaining for them). You again give one tennis ball to the dog after all four people have done the same and divide the remaining tennis balls evenly among all 4 people and you. There are no more balls left. What is the minimum number of tennis balls that could have been in the center of the room at the beginning?

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marked as duplicate by grand_chat, Leucippus, eranreches, Namaste algebra-precalculus Jan 5 '18 at 0:51

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The (in)famous solution due to Dirac

We have $-4$ balls in a room. (We, the company, owe the next door pimp $4$ balls.)

1 You give one ball to the dog. (You borrow one more ball so that you can give it to the dog.) There remains $-5$ balls in the room. (The company owes the next door pimp $5$ balls now.) You divide the balls into $5$ equal groups, and take away one group containing exactly $-1$ balls. (You agree with the pimp that you will take care of the fifth ball borrowed.) There remain $4$ such groups. Then you merge these groups and there remains again $-4$ balls. (The company owes the next door pimp $4$ balls.)

2 ... 4

The company still has $-4$ balls. (The company still owes the next door pimp $4$ balls.) You give one ball to the dog and each of you takes away $-1$ balls. (Each of you agrees with the pimp to take care of one ball personally.)

There remains $0$ ball. (No further common liability to the next door pimp.)

So, if the question would have been about the minimal absolute value of the number of balls to start with then the answer would have been $$4.$$

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  • $\begingroup$ O! cruel, cruel Exchange wherein I can upvote this but once. $\endgroup$ – B. Goddard Jan 4 '18 at 23:22

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