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"Orange juice, please" I said to my new friend after I told her my solution, "with a touch of vodka."
Solutions:
1. Because of Ron, there is nobody who has shaken hands with 8 others.
So, the numbers are:
0, 1, 2, 3, 4, 5, 6, 7, X, where X is the unknown number.
Every time two people shake hands, we’ll say that each of the two has done a "half-shake." Because each act of shaking hands increases the total count of all "half-shakes" by two and there is an even number of half-shakes without the X, X must be an even number.
Could X be 6? That is, could the shakes be:
0, 1, 2, 3, 4, 5, 6, 7, 6
No, because 7 (we'll identify people with the number of half-shakes they do) must shake hands with everyone (including 1) and shake hands with 2. This eliminates 1 from shaking hands with either of the 6s. That implies that both 6s would have to shake hands with 2, but then the 2 would become a 3.
By contrast, the following is possible and is the best we can do:
0, 1, 2, 3, 4, 4, 5, 6, 7
7 shakes with 1, 2, 3, 4, 4, 5, 6.
6 shakes with 2, 3, 4, 4, 5, 7.
So, 1 and 2 are done.
5 shakes with 3, 4, 4, 6, 7.
Now 3 is done.
The 4s shake with one another.
Is 0, 1, 2, 2, 3, 4, 5, 6, 7 possible?
No.
7 shakes with 1, 2, 2, 3, 4, 5, 6.
6 shakes with 2, 2, 3, 4, 5, 7.
So, both 2s are done.
This does not leave people for the 5s.
2. Before the last handshake, here was the situation:
0, 1, 2, 3, 4, 4, 5, 6, 7
Notice that everyone who had shaken at least 4 hands must have shaken hands with all others who had shaken at least 4 hands. That is, the 4s, 5s, 6s, and 7s had all shaken hands with one another. Therefore neither Caroline nor I could be among those, because we first shook hands after I resolved the second problem. Therefore, we must be among the 1, 2, 3. But Caroline shook more hands than I did, but I do not know how many unless I had shaken two hands before shaking hers. In that case, she has now shaken four hands.
