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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: 7,659 Thanks: 2635 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: 14,415 Thanks: 1025
June 6th, 2014, 12:42 PM   #6
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Quote:
 Originally Posted by Shen 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
Total even hand shake = EH+NH (0 is an even number)

June 6th, 2014, 01:14 PM   #7
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Quote:
 Originally Posted by mathman (0 is an even number)
Even so...

June 6th, 2014, 02:59 PM   #8
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Quote:
 Originally Posted by Shen 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.
It doesn't matter how many people there are. But let's take 100. Then

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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### prove that the number of people who shake hands of an odd number of people is always even

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