New Scientist wrote: A married couple, Noel and Holly, invite four other couples over for drinks. It's a rather formal affair, but as everyone in the group is a mathematician they don't want to do more handshaking than is necessary. Each person only shakes hands with the people they haven't met before.
As the brandy begins to flow, Noel asks everyone at the party how many hands they shook, and he receives nine different answers. How many people did Holly shake hands with?
Handshakes
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Handshakes
I enjoyed this puzzle in the Christmas edition of New Scientist. I'm pretty sure I have the answer right, but it won't be published till next year...
StuartR
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Re: Handshakes
Spoiler
There may be maths involved to calculate the answer, but my gut says that none of the answers can be verified if the brandy was already flowing freely!
Regards,
Rudi
If your absence does not affect them, your presence didn't matter.
Rudi
If your absence does not affect them, your presence didn't matter.
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Re: Handshakes
Spoiler
None. As joint host, she would know everyone already. But I'm not certain
Leif
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Re: Handshakes
Spoiler
My answer is that Holly shook hands with 4 people, just like Noel himself.
Although mathematicians can be rather autistic (I am one myself ), I assume that they know their own partner, so the maximum number of handshakes per person is 8, and the minimum is 0 of course.
Since Noel hears 9 different answers, he must have heard each of 0, 1, 2, ..., 8 once.
One person has shaken 8 hands, i.e. everyone except his/her partner. As a consequence, all those others have shaken at least 1 hand, and hence the partner of the first person must be the one who has shaken none.
Then, there is someone who has shaken 7 hands: everyone except the partner of the first person and his/her own partner. These 7 have shaken at least 2 hands, so the partner of this second person must be the one who has shaken only 1 hand.
Continuing this way, we have a couple of which one has shaken 6 hands, and the other only 2.
Then a couple of which one has shaken 5 hands, and the other 3.
In the final couple, both partners have shaken 4 hands. This MUST be Holly and Noel, otherwise Noel would have received 4 twice as an answer.
Although mathematicians can be rather autistic (I am one myself ), I assume that they know their own partner, so the maximum number of handshakes per person is 8, and the minimum is 0 of course.
Since Noel hears 9 different answers, he must have heard each of 0, 1, 2, ..., 8 once.
One person has shaken 8 hands, i.e. everyone except his/her partner. As a consequence, all those others have shaken at least 1 hand, and hence the partner of the first person must be the one who has shaken none.
Then, there is someone who has shaken 7 hands: everyone except the partner of the first person and his/her own partner. These 7 have shaken at least 2 hands, so the partner of this second person must be the one who has shaken only 1 hand.
Continuing this way, we have a couple of which one has shaken 6 hands, and the other only 2.
Then a couple of which one has shaken 5 hands, and the other 3.
In the final couple, both partners have shaken 4 hands. This MUST be Holly and Noel, otherwise Noel would have received 4 twice as an answer.
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Best wishes,
Hans
Hans
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Re: Handshakes
Hans, that is exactly the answer that I came up with, and following the same logic.
StuartR
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Re: Handshakes
I think it's a delightful puzzle because at first it seems as though there can't be enough information
StuartR
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Re: Handshakes
I'm still more interested in the brandy? Any calculation to determine what brand of brandy was flowing?
Regards,
Rudi
If your absence does not affect them, your presence didn't matter.
Rudi
If your absence does not affect them, your presence didn't matter.
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Re: Handshakes
The nutters on my Facebook page kept making all sorts of weird suggestions for this one, so I had to post:
Can anyone here think of any more assumptions that I might have forgotten to document for them?Stuart's Facebook page wrote: OK folks. Since so many of you are struggling. Here are the assumptions that I am aware of having made:
• Every couple consists of two people who are capable of shaking hands
• The two people in each couple have met each other, and therefore do not shake hands
• Each handshake involves exactly two people, never more or less
• Every handshake is transitive. If A shakes hands with B, then B shakes hands with A
• Each person correctly reports the number of handshakes they have been involved in
StuartR
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Re: Handshakes
I totally agree with Rudi ! You have my vote big guy!Rudi wrote:I'm still more interested in the brandy? Any calculation to determine what brand of brandy was flowing?
Regards,
Bob
Bob
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Re: Handshakes
On a similar, but different topic...
Q: How many mathematicians does it take to screw in a light bulb?
A: It only takes one mathematician to screw in a light bulb.
Proof:
[/size]Source
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Regards,
Rudi
If your absence does not affect them, your presence didn't matter.
Rudi
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Re: Handshakes
Most excellent proof, Rudi!
Bob's yer Uncle
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