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**4 members have voted**

December 3rd, 2017 at 9:53:43 PM
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At the 2018 WoV Spring Fling there were many handshakes. The total number of people who shook hands an odd number of times was even. Prove this to be true.

Free beer to the first acceptable proof.

Free beer to the first acceptable proof.

It's not whether you win or lose; it's whether or not you had a good bet.

December 3rd, 2017 at 10:22:25 PM
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There were 4 people. They all shook hands with every other person, once each. So 4 people shook 3 hands each.

#Easy

#Actually,SeemsTooEasy

#Meh

#ShipTheSherbert

#Covfefe

#Easy

#Actually,SeemsTooEasy

#Meh

#ShipTheSherbert

#Covfefe

It's a piece of cake to bake a pretty cake. If the way is hazy. You gotta do the cookin' by the book. You know you can't be lazy. Never use a messy recipe. The cake will end up crazy. If you do the cookin' by the book. Then you'll have a...
Break it down! Lemme see you bake it up! Drop that down low and pick that up. Break it down! Lemme see you bake it up! Drop that down low and pick that up Now back that Hey! Now back that. Hey! Now bake that. Hey! Now bake that. Hey!
It's a piece of cake to bake a pretty cake (WHAT!) If the way is hazy (OKAY!) You gotta do the cookin' by the book, (WHAT!) You know you can't be lazy (YEAH!) Never use a messy recipe, (WHAT!) The cake will end up crazy (OKAY!) If you do the cookin' by the book, (YEEAAHHHH!) Then you'll have a...cake!
Rub that, it's yours! Grab this, it's yours! Rub that, it's yours! Grab this, it's yours! Now turn around, put that on. Grind, make it get a little bigger! Now turn around, put that on.
Grind, make it get a little bigger!
You gotta do the cookin' by the book! HEY!

December 3rd, 2017 at 10:35:13 PM
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Quote:WizardAt the 2018 WoV Spring Fling there were many handshakes.

If you can beat games using time travel, is that AP or cheating?

"You can tell it's real cause it looks so fake" - Musk

December 3rd, 2017 at 11:12:26 PM
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Quote:RogerKintQuote:WizardAt the 2018 WoV Spring Fling there were many handshakes.

If you can beat games using time travel, is that AP or cheating?

Well we know what he did with the proceeds form the sale of the site,

he went out and bought himself one of those Deloreans with the flux capacitor.

December 4th, 2017 at 6:21:25 AM
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Quote:RogerKintQuote:WizardAt the 2018 WoV Spring Fling there were many handshakes.

If you can beat games using time travel, is that AP or cheating?

It has happened already with disastrous results

https://wizardofvegas.com/forum/off-topic/general/646-a-fool-proof-system-to-beat-roulette/

December 4th, 2017 at 6:33:04 AM
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Quote:RSThere were 4 people. They all shook hands with every other person, once each. So 4 people shook 3 hands each.

Okay, there's one situation where it works but you have to show it is true of every situation.

#NoBeerForYou

It's not whether you win or lose; it's whether or not you had a good bet.

December 4th, 2017 at 7:32:59 AM
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I am a physicist not a mathematician, which must be why I have just written the must garbled and poorly-written mathematical proof in the history of humanity.

Even-ness can be defined as being congruent with 0mod(2), odd-ness can be defined as congruence with 1mod(2)

Every handshake involves two "half-shakes" -one "half-shake" per person. Therefore, no matter how many handshakes there are, the total number of "half-shakes" must be an even number because

if n = total number of handshakes, then the total number of 'half-shakes' =2*n, and 2*n must be an even number, mathematically 2n | 0mod(2) (2n is congruent with 0mod(2) )

Note that when each nerd at the event shakes hands, they are only counting their own "half-shake"

Now, let k =number of people who shook hands an even number of times, 0mod(2)

m=number of people who shook hands an odd number of times, 1mod(2)

the condition that must be satisfied, because 2n | 0mod(2), is:

0mod(2) = k * 0mod(2)+m * 1mod(2) (Eq. 1)

No matter whether k is even or odd, the expression (k * 0mod(2) ) must be even, i.e., (k * 0mod(2) ) | 0mod(2), . So this means that the second term ( m * 1mod(2) ) must also be even, i.e. it must congruent with 0mod(2) to satisfy Equation 1. The only way for ( m * 1mod(2) ) to be even is if m is even, or if m | 0mod(2).

Thus m, the number of people having an odd number of handshakes, must be even.

In English

if you ask every person how many handshakes they experienced, the sum of their answers must be even (because there are two people for each handshake.). Now separate all the people into two groups; Group 1: those who had an odd number of handshakes and Group 2: those who had an even number of handshakes.

In Group 2, the sum of the handshakes will always be even, because any number of people, whether odd or even, times an even number of handshakes will yield a total number that is an even number.

Thus, the total number of handshakes in group 1 must also be even. Since everyone in group 1 had an odd number of handshakes, the only way the sum of their handshakes can be even is if there is an even number of people in Group 1.

Every handshake involves two "half-shakes" -one "half-shake" per person. Therefore, no matter how many handshakes there are, the total number of "half-shakes" must be an even number because

if n = total number of handshakes, then the total number of 'half-shakes' =2*n, and 2*n must be an even number, mathematically 2n | 0mod(2) (2n is congruent with 0mod(2) )

Note that when each nerd at the event shakes hands, they are only counting their own "half-shake"

Now, let k =number of people who shook hands an even number of times, 0mod(2)

m=number of people who shook hands an odd number of times, 1mod(2)

the condition that must be satisfied, because 2n | 0mod(2), is:

0mod(2) = k * 0mod(2)+m * 1mod(2) (Eq. 1)

No matter whether k is even or odd, the expression (k * 0mod(2) ) must be even, i.e., (k * 0mod(2) ) | 0mod(2), . So this means that the second term ( m * 1mod(2) ) must also be even, i.e. it must congruent with 0mod(2) to satisfy Equation 1. The only way for ( m * 1mod(2) ) to be even is if m is even, or if m | 0mod(2).

Thus m, the number of people having an odd number of handshakes, must be even.

In English

if you ask every person how many handshakes they experienced, the sum of their answers must be even (because there are two people for each handshake.). Now separate all the people into two groups; Group 1: those who had an odd number of handshakes and Group 2: those who had an even number of handshakes.

In Group 2, the sum of the handshakes will always be even, because any number of people, whether odd or even, times an even number of handshakes will yield a total number that is an even number.

Thus, the total number of handshakes in group 1 must also be even. Since everyone in group 1 had an odd number of handshakes, the only way the sum of their handshakes can be even is if there is an even number of people in Group 1.

December 4th, 2017 at 8:49:55 AM
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I think this has been explained already, but I'll try to do a simple one.

If there are N people, each person will shake hands N-1 times to shake hands with everybody.

In order to shake hands an odd number of times, N must be an even number, so that N -1 (total handshakes per person) is odd.

Since N is an even number, and for every person in N there are an odd number of handshakes, the number of people with an odd number of handshakes is also N, which is an even number.

If there are N people, each person will shake hands N-1 times to shake hands with everybody.

In order to shake hands an odd number of times, N must be an even number, so that N -1 (total handshakes per person) is odd.

Since N is an even number, and for every person in N there are an odd number of handshakes, the number of people with an odd number of handshakes is also N, which is an even number.

December 4th, 2017 at 8:56:26 AM
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Quote:gordonm888I am a physicist not a mathematician, which must be why I have just written the must garbled and poorly-written mathematical proof in the history of humanity.

I believe your proof only works in a perfect vacuum.

My goal of being well informed conflicts with my goal of remaining sane.

December 4th, 2017 at 9:07:31 AM
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Let N be # of people where N = a non-negative integer. N people shake hands with N-1 people.

If N is odd then N-1 is even. Thus 0 people shake hands an odd # of times. Zero is an EVEN number

If N is even then N-1 is odd. Thus in either case the assumption is true.

My goal of being well informed conflicts with my goal of remaining sane.