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let's say we live in a "very strange" world of Facebook, in which everyone has $1000$ friends. In addition, every $n$ people will have exactly $\lceil{1000 * (\frac{1}{10})^{n-1}}\rceil$ friends in common if $n \leq 3$, not more than a friend if $n$ between (inclusive) $4$ and $1000$ and $0$ otherwise. How many people are there in this world? If there can be more than an answer, what is the lower and upper bound for this number?

PS: Friends and self are excluded from friends of friends.

PS2: I don't know the answer either :)

PS3: Assumed that there must be at least a person in this world

Edited:Just found a "bug" for this problem, say I'm A and my friends are A1, ..., A1000, friend in common of A, A1, A2, A3, A4 must be subset of friend in common of A1, A2, A3, A4, which is $A$.

I need to change the statement of the problem... :(

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    Did you make this up? The rounding up does not help.2011-12-06
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    The square parenthesis mean the floor function? In that case it is not enough to restrict to $n\leq 4$?2011-12-06
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    @Henry: yes I made this up... Well it might not be very helpful, but without it, we may have a non-integer number of friend in common...2011-12-06
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    @Student73: no, it was ceiling, actually the ones important is only n=1,2,3, but if I am A and my friends are A1, ..., A1000, there must be friend of common for A1, A2, ..., A1000, which is A, myself :)2011-12-06
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    You are right! :)2011-12-06
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    Is there a reason to think this state of affairs is even possible?2011-12-06
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    @mjqxxxx: it's a riddle, everything can happen in a riddle :)2011-12-06
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    Quick question - is *Kevin Bacon* in this world?2011-12-06
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    So, every 1000 people have exactly 1 friend in common. Every 1001 people have exactly 0 friends in common.2011-12-06
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    @Graphth: that was the idea...2011-12-06
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    bug found (written in the question), if someone can restate the problem (that's still same with my idea), please comment2011-12-06
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    @TheChaz: I don't get your question... Is it some kind of joke?2011-12-06
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    Yes, it is a joke, but refers to (what could be) a related riddle about degrees of separation between circles of friends.2011-12-06
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    And, this is why you don't make up random problems. But, what you said doesn't make sense. You haven't told us the friends of $A_1$, $A_2$, $A_3$, $A_4$, so we don't know who the common friends of those 4 are. It is true that the common friends of those and $A$ must be a subset of $\{A_1, \ldots, A_{1000}\}$. But, if $A$ is not included, the friends of $A$ have little to do with it. I'm sure there is some bug in this problem, but I don't want to figure it out.2011-12-06
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    I think it's an interesting question - er - riddle. Reminds me of the research that tried to demonstrate that in a chain of friends there are 6 persons on average in the entire world. So if John Doe wants to pass a message to Obama, the message will on avg pass through 4 more people (if John Doe is not a personal acquaintance of Obama that is...)2011-12-06
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    You ought to close this problem, or add a comment to the top at least, so that people don't come here and read through your question only to find you've essentially un-asked it.2011-12-06
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    @ThomasAndrews: done...2011-12-06
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    @zfm: How about this: each person has $1000$ friends; each pair of people has $100$ friends in common; each triplet of people has $10$ friends in common; and every $k$-tuple of people with $k>3$ has *at most* $1$ friend in common.2011-12-06
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    @mjqxxxx: very interesting idea! on the other hand, will we have one answer or the answer will be bounded?2011-12-06

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