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10 people can share a bucket of coins equally. A monkey steals one coin. The number of coins remaining is therefore 1 less than would allow them to be shared out equally.

One person after another tries to take the coin from the monkey -- but each time, the monkey kills them. Each time a person dies, the number of coins is still 1 less than would allow the remaining people to share them out equally, until we are left with just one person who wisely leaves the monkey alone and takes away ... how many coins?

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marked as duplicate by JMP, Gamow, Glorfindel, Beastly Gerbil, Mike Earnest Jul 24 '17 at 18:32

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This is very similar to, for instance, this puzzle: N pirates steal their share of bananas to the benefit of a monkey.

We're looking for a number that's 1 less than a multiple of 2,3,4,...,10. In other words, when we add 1 we get a number that's a multiple of all those numbers. Equivalently, that's a multiple of their LCM, which is 8x9x5x7=2520.

Therefore

the number of coins in the bucket was one less than some (positive) multiple of 2520. The smallest possible number is 2519.

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The answer is

# of coins = $2519+2520x$.

so the minimum number of coins is

$2519$

I have used a simple code you can run by yourself: http://rextester.com/AKQ5063

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