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June 5th, 2014, 07:35 PM  #1 
Member Joined: Apr 2014 From: Missouri Posts: 39 Thanks: 1  Hand shaking problem
Hi I did this problem in my way but i get different result. A group of 25 people are in a party. Many people shake hand with one another. Let n be the number of people who shook hand with odd number of people. Prove that n is even. Prove that n is even if there are 100 people in that party I did it in this way. 25= EH+OH+NH where EH= Even hand shaken people, OH= Odd hand shaken people and NH= No hand shaken people now if NH= even then EH= even and if NH = odd then EH= odd. Now according to my doing n could be even or odd. Whats the problem here? Thanks in advance 
June 5th, 2014, 08:45 PM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,445 Thanks: 2499 Math Focus: Mainly analysis and algebra 
If you have an odd number of people that shook hands with an odd number of people, you have half a handshake. Or, to state it another way, there are two people involved in every handshake, so the total number of handshakes increases by two each time. And the total is 0 x NH + 2p x EH + (2q + 1) x OH for some integers p, q. This total is odd if and only if OH is odd, hence OH is even. 
June 6th, 2014, 08:10 AM  #3 
Member Joined: Apr 2014 From: Missouri Posts: 39 Thanks: 1 
Hi V8archie How we are sure that there are 0 people involved in NH? How odd number of people if shake hand with odd number of people then we will have half a hand shake? Like if 3 people hand shook with 3 people we will have 15 hand shake. 0*NH+2p*EH+(2q+1)*OH= total hand shake = 300 so it is even. The result alters then. I am little bit confused. Thanks 
June 6th, 2014, 08:25 AM  #4 
Member Joined: Apr 2014 From: Missouri Posts: 39 Thanks: 1 
Hi V8archie If there are 100 people then total is even. then EH needs to be odd then. and then there is no valid solution as well. Like 100= 2p*EH+(2q+1)*OH, if 2p is even then first term is always even. we need even in the second term. but (2q+1) is odd and only give even when OH is even but our assumption states OH to be odd. Thanks...... 
June 6th, 2014, 09:48 AM  #5 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,282 Thanks: 931  
June 6th, 2014, 12:42 PM  #6  
Global Moderator Joined: May 2007 Posts: 6,607 Thanks: 616  Quote:
 
June 6th, 2014, 01:14 PM  #7 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,282 Thanks: 931  
June 6th, 2014, 02:59 PM  #8  
Math Team Joined: Dec 2013 From: Colombia Posts: 7,445 Thanks: 2499 Math Focus: Mainly analysis and algebra  Quote:
100 = EH + OH + NH We don't know how many handshakes there were (and we don''t care), but let's call it H. Then 2H = 2p EH + (2q + 1) OH 2H, because we are counting both sides of each handshake (person handshakes if you like), is even whatever the value of H. 2p EH is also even whatever the values of p and EH. So (2q + 1) OH must be even, and since (2q + 1) is always odd, whatever the value of q, OH must therefore be even. Given our 100 people, if EH is odd, then NH must also be odd. EH + NH is always even, there's nothing you can do about it. If the number of people is odd, EH + NH is odd.  

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