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.
On
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.
On
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.
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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
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.