I just bought The USSR Olympiad Problem Book; get ready for some difficult math problems in the coming days, @Kainui , @ganeshie8 , @dan815 ,@iambatman , @Jhannybean , @perl . @Zarkon .

Question

mendicant_bias · Answer

Let's take a shot at number one:

"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 a even number."

jtvatsim · Answer

A handshake happens between a pair of individuals. No matter the criteria, the number of people who have shaken hands (assuming that they didn't do so with themselves) must be even.

mendicant_bias · Answer

(Looks at book solution)

jtvatsim · Answer

Actually, let me modify that answer... there's a minor technicality that I need to sort out.

jtvatsim · Answer

The spirit of the argument is what I meant though. :)

mendicant_bias · Answer

"Consider the total number of handshakes which have been completed at any moment. This must e an even number, since every handshake is participated in by two people, thus the total number is increased by two. The number of handshakes, however, is also the sum of the handshakes made by each individual person. Since this sum is an even number, the count of the people who have shaken hands an odd number of times must be even (otherwise, odd times odd would given an odd contribution to the total.)"

jtvatsim · Answer

Yeah, what they said! ;)

jtvatsim · Answer

The criteria that they all be people who shook hands an odd number of times DOES matter.

mendicant_bias · Answer

This is Problem 1/320. It later gets into complex numbers, but is mostly actually basic arithmetic/algebra stuff, but just extremely complicated versions of it. Most people on this website should be able to solve every single one....we'll all just have a hard time with most of them, lol. 

You know what would be a cool challenge? OpenStudy solving all 320 problems collectively in 30 days, or something, call it "The Olympiad Challenge" and try to finish it over most of our school breaks.

jtvatsim · Answer

That would be intense! XD

mendicant_bias · Answer

Indeed. And thus I declare it so. I'm going to read over that answer again and post the second one now, heh.

ganeshie8 · Answer

Hey looks i  did not follow the question/solution completely 

number of handshakes  = $\large \binom{n}{2}$
how is this an even number always ?

mendicant_bias · Answer

I'm still looking at it. It makes sense to me that it's always an even number, though, why wouldn't it be?

ganeshie8 · Answer

Gotcha! they did not shake hands with everyone, they only `shaken hands with a certain number of other persons. `

mendicant_bias · Answer

Since odd*even=even, even*even=even, even*odd=even, and only odd*odd = odd, you already know that one factor you are multiplying by, the total number of handshakes, is even; thus, assuming for the sake of argument that you are only referring to those who have shaken hands an odd number of times, 

yuh

jtvatsim · Answer

Yeah, that is it.

jtvatsim · Answer

The criteria that matters is that they did so within a group of people following the same rule (shaking hands with an odd number of people).