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This is from a book called USSR Olympiad Problem book:

Every living person has shaken hands with a certain number of other persons. Prove that a count of the number of people who have shaken hands an odd number of times must yield an even number.

I've got no clue how to even start tackling the problem. I've checked the hints page still don't understand.

Would be nice if you would show me the logic behind this.

Did
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Evgeny Danilenko
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  • What does the hints page say? – Did Dec 14 '13 at 20:27
  • First show that the total of all handshakes made up to any time is an even number. – Evgeny Danilenko Dec 14 '13 at 20:32
  • Depends on the definition of "the number of handshakes": with one definition this is trivially wrong and with the other this is trivially right, won't you say? And what is your definition of "the number of handshakes"? – Did Dec 14 '13 at 20:36
  • Can someone like, help me solve it? I came here for help - thanks for the answers and everything but I don't understand the "math" talk. – Evgeny Danilenko Dec 14 '13 at 21:05

4 Answers4

3

Each handshake involves 2 people. If the $i$th person shakes hands $n_i$ times, then the sum of the $n_i$ must be even. If there were an odd number of odd $n_i$ then the sum would be odd.

guest196883
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  • I've tried understanding your answer but can you please simplify it further? I'm a noob in maths. started taking in seriously pretty recently. after I failed aptitude test for a software company :'( – Evgeny Danilenko Dec 14 '13 at 19:50
  • Sorry Evgeny, but which part exactly do you want to see simplified?? – Did Dec 14 '13 at 20:25
  • "If the ith person shakes hands ni times, then the sum of the ni must be even. If there were an odd number of odd ni then the sum would be odd." – Evgeny Danilenko Dec 14 '13 at 20:27
  • Make up your mind: first quoted sentence or second quoted sentence? – Did Dec 14 '13 at 20:37
  • all of which I have quoted, is what I didn't understand. translate it to english lol – Evgeny Danilenko Dec 14 '13 at 20:47
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    So the total number of hands shaken has to be even, because each handshake involves two people. Do you understand this part? – guest196883 Dec 14 '13 at 21:15
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Every time you shake hands you get a euro. An even number of euro are distributed. Those with an even number of handshakes between them have an even number of euro So those that have an odd number of handshakes must between them have an even number of euro This can only happen if there are an even number of them

Brian Woods
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All people here have only one hand (that is one confusion kicked out!). Draw a circle and place numbers $1,2,3,\dots$ spaced equally on its circumference; to represent the single hands of persons. Connect the numbers as you wish; each connection representing a handshake. Example: If $4$ lines come out from say Hand $5$, then Person $5$ has had $4$ handshakes; $4$ being the degree of Hand $5$. Find the degrees of all the hands. Some degrees will be odd; some even. The number of odd degree hands will always be even. Why? Figure it out. Not difficult.

Mohsen Shahriari
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Subramanyan R
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The trouble with this is, some living people have shaken hands with people who are now dead. If you want this to work, you have to either count the dead people as well as the living, or only count handshakes with people who are still living.

Robert Israel
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