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Dr. and Mrs. Smith recently attended a party at which there were 5 other couples. Various handshakes took place. No one shook hands with his/her spouse, no one shook his/her own hand. After all the handshaking was finished, Mrs. Smith asked each person, including her husband, how many hands he or she had shaken. To her surprise, each person gave her a different answer.

How many hands did Dr. Smith shake?

Note: I don't actually know the answer to this question, so I'll accept the most plausible one.

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  • $\begingroup$ I guess we can assume that Dr. and Mrs. Smith are spouses to each other (and not just randomly having the same name), and only right hands are shaken? $\endgroup$ – Paŭlo Ebermann Mar 31 '15 at 16:12
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    $\begingroup$ Welcome to the site, Narida! From what I've seen on meta, the goal of this site is to collect original problems only, or questions to which answers could not be found on a simple search. $\endgroup$ – Aravind Mar 31 '15 at 19:48
  • $\begingroup$ Actually, this should be "Dr. and Mrs Owe", as the puzzle was created by Lars Bertil Owe. (In the original - at least as reproduced in Sci. Am.) Lars himself asked the questions and wanted to know how many hands his wife shook. And there were 5 couples total. $\endgroup$ – Paul Sinclair May 13 '16 at 23:42
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EDIT: I misread the question. There are 5 other couples, not 5 couples total.


I'm going to assume that you're asking how many distinct people whose hands were shook. Shaking the other hand of the same person or the same hand twice still only counts as one. That is a necessary constraint for a unique answer. In that case, Mr Smith shook

5 hands

Because

Among the five married couples no one shook more than ten hands. Therefore, if the eleven people asked by Mrs. Smith each shake a different number of hands, the numbers must be 0, 1, 2, ..., 10. The person who shook 10 hands has to be married to the person who shook 0 hands (otherwise that person could have shaken only nine hands.) Similarly, the person who shook 9 hands is bound to be married to the person who shook 1 hand. So that the married couples shook hands in pairs 10/0, 9/1, 8/2, 7/3, 6/4. The only person left who shook hands with 5 is Mrs. Smith. Therefore, Mr. Smith shook the pair number for 5 which is 5.

To help clarify:

Since there are an even number of people and they all shook a different number, there was going to be a pair in the middle that shook the same number hands. The only way that is possible is if Mr. Smith - the only one not included in the "different number" condition - shook the same number of hands as someone else.

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  • $\begingroup$ Nice answer, but there were actually 6 couples: Smiths + 5 others. I think your approach still works with 10/0, 9/1, ..., 5/5. Mr Smith then shook 5 hands. $\endgroup$ – Lawrence Mar 31 '15 at 12:35
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I'd like to start by stating that "Engineer Toast" has the most reasonable solution. However, his answer relies on assumptions and it is actually unclear (semantically speaking) how many hands Dr. Smith shook.

All that I can conclude is that Dr Smith shook between a lower bound of 0 and an upper bound of 20 hands.

Reason:

There are 6 couples total (12 people)
There are most likely 2 hands per person (24 total hands)
Dr. Smith cannot shake his or his wife's hands leaving up to 20 hands that he can shake.
Since there are a number of ways that could play out it is unclear how many hands he actually shook.

An even more ridiculous answer (if you hated my first answer look away now):

Depending on the type of party, each party-goer may have had too much to drink and made up a number because they really had no clue. In other words: Just because they gave an answer, it doesn't mean the answers they gave were correct.

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  • $\begingroup$ So I realize full well that this isn't a "great" answer, but I'd still like to know reason for the down vote. Is there something that I said that was incorrect? $\endgroup$ – Warlord 099 Mar 31 '15 at 19:51
  • $\begingroup$ +1 for pointing out that each person has more than one hand. However, this, too, makes assumptions. Due to mutations or accidents, there could be many more or less hands than 24. The question should be phrased to require counting people with whom you have shook hands as that number is known. For that matter, it should state that there were exactly 5 other couples and nobody else. What about single people? It sounds like a question best asked to people who will make the common assumptions instead of a puzzler looking for loopholes and snarky answers. $\endgroup$ – Engineer Toast Mar 31 '15 at 20:01
  • $\begingroup$ @Warlord099, my mistake to think that you were just nitpicking; you are correct to point out that the question is not phrased precisely - even the possibility that one pair of persons may have shaken hands several times,is not explicitly forbidden. $\endgroup$ – Aravind Mar 31 '15 at 20:19
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    $\begingroup$ @EngineerToast I agree 100% with your last statement which is why I believe 100% you have the best answer. The main reason I brought up the "multiple hands loophole" is because the "shake your own hand loophole" was made explicitly illegal (the only reason I posted the bottom part is I was already posting and was trying to be funny). And although I agree he left it open by not saying there were no single people involved, it is much more of a stretch to make up characters for a problem than it is to assume most people have two hands ;) $\endgroup$ – Warlord 099 Mar 31 '15 at 20:28
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Also assume this arrangement for party members

1)Dr. Smith<---do handshake with-->Mrs.Smith,Her Husband,C1,c1,C2,c2,C3,c3,C4,c4,C5,c5

2)Mrs. Smith and her husband

3)C1 c1

4)C2 c2

5)C3 c3

6)C4 c4

7)C5 c5

Where c1,c2,... denotes 5 other couples...

Then in this case Dr. Smith shook 12 Hands distinctly.

Because its not mentioned that in party all were couples.

So there are 12 handshakes where it will satisfy all the rules/cases for handshakes that are possible for Dr. Smith.

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  • $\begingroup$ If it said "Dr Smith and Mrs Smith" I'd probably agree with you, but since it said "Dr and Mrs Smith" I think they are quite obviously together... $\endgroup$ – Warlord 099 Mar 31 '15 at 19:33
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There's not enough information in the question to be able to answer. If Dr. Smith shook ALL available hands, the answer is 10 (5 couples).

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  • $\begingroup$ Look at the other answer, surprisingly there is enough information! $\endgroup$ – Falco Mar 31 '15 at 17:49

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