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.
Answer
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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