Donkey Kong Forum
Other Classic Arcade Games => Classic Arcade Game Discussion => Topic started by: Weehawk on August 31, 2014, 03:18:10 pm
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(http://www.slither-gdi.net/crams.jpg)
Jason and Anna Cram went to an arcade party where they and four other married couples were the only people present.
Some of the people at the party had met before, some had not. During the course of the evening, each pair of people that had not met previously made it a point to play a game of Donkey Kong together. No other Donkey Kong was played.
At the end of the party, Jason asked everyone in attendance, including Anna, how many people they played Donkey Kong with. To his surprise, Jason got a different answer from each person.
How many people did Anna play Donkey Kong with at the party?
For the first person who can post the correct answer with an explanation why it is correct who does not already have one, I have a t-shirt:
(http://www.slither-gdi.net/2014shirt.jpg)
Please recuse yourself if you were already familiar with the problem.
If you already have the shirt, you can PM your answer to me and I will credit you in the thread.
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Are you positive that there isn't a piece of information missing here? ;D
"Someâ„¢" had or had not met before...
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<mad>
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6?
Edit: Too stupid to see the explanation thing. Suck!
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Here's what I have so far and I don't even know if I'm on the right track at all, so somebody can take this and run with it, fttt this I'm gone.
- There are 10 people total
- The maximum number of people each individual could play DK with is 8 (since they would definitely know themselves and their spouse, but could potentially not have met any of the others).
- The question says that *each* person gave Jason a different answer. I interpret that to mean that he got 9 *distinct* answers (that is, nobody's answer was the same as anybody else's answer).
- There are 9 possible answers, zero through 8.
- However, zero AND 8 could not have BOTH been given as answers. One or the other is possible, but not both. If an attendee played zero games, it means they knew everybody at the party, which means that there could not have been an attendee who played 8 games, because you can't simultaneously have a person who has met everybody and a person who has met nobody.
- So Jason only could have gotten 8 distinct answers.
That's where I'm stuck, and I haven't even gotten to how Anna's answer could be narrowed down from the information given.
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Here's what I have so far and I don't even know if I'm on the right track at all, so somebody can take this and run with it, fttt this I'm gone.
- There are 10 people total
- The maximum number of people each individual could play DK with is 8 (since they would definitely know themselves and their spouse, but could potentially not have met any of the others).
- The question says that *each* person gave Jason a different answer. I interpret that to mean that he got 9 *distinct* answers (that is, nobody's answer was the same as anybody else's answer).
- There are 9 possible answers, zero through 8.
- However, zero AND 8 could not have BOTH been given as answers. One or the other is possible, but not both. If an attendee played zero games, it means they knew everybody at the party, which means that there could not have been an attendee who played 8 games, because you can't simultaneously have a person who has met everybody and a person who has met nobody.
- So Jason only could have gotten 8 distinct answers.
That's where I'm stuck, and I haven't even gotten to how Anna's answer could be narrowed down from the information given.
What you said, and my answer is 7
since 0 and 8 can't be given as answers. ;D
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Chris, we seriously have to stop meeting like this FailFish I'll type faster next time...
If I'm reading this correctly...
Jason asked everyone in attendance how many people they played Donkey Kong with, and got a different answer from every person.
There were 10 people in total at the event, or 5 couples, which means the most number of different people any one person could have played with is 8 (because you can't play with yourself or your partner).
Thus, for Jason to have received 9 different answers, those answers must have been 8,7,6,5,4,3,2,1, and 0.
However, if someone played with 0 other people, then suddenly the maximum number of different people any one player could have played with becomes 7, and the problem no longer makes sense.
So, what am I misinterpreting about the Arcade House Party Conundrum?
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Here's what I have so far and I don't even know if I'm on the right track at all, so somebody can take this and run with it, fttt this I'm gone.
- There are 10 people total
- The maximum number of people each individual could play DK with is 8 (since they would definitely know themselves and their spouse, but could potentially not have met any of the others).
- The question says that *each* person gave Jason a different answer. I interpret that to mean that he got 9 *distinct* answers (that is, nobody's answer was the same as anybody else's answer).
- There are 9 possible answers, zero through 8.
- However, zero AND 8 could not have BOTH been given as answers. One or the other is possible, but not both. If an attendee played zero games, it means they knew everybody at the party, which means that there could not have been an attendee who played 8 games, because you can't simultaneously have a person who has met everybody and a person who has met nobody.
- So Jason only could have gotten 8 distinct answers.
That's where I'm stuck, and I haven't even gotten to how Anna's answer could be narrowed down from the information given.
What you said, and my answer is 7
since 0 and 8 can't be given as answers. ;D
The answer is incorrect, and the reasoning you borrowed is fatally flawed.
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It's a classic arcade party so obviously there's a few liars and cheats in there, it won't add up.
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ROFL
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It's a classic arcade party so obviously there's a few liars and cheats in there, it won't add up.
I naively overlooked this. In my fictional scenario everyone told the truth.
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I've got it!
One of the party goers was a schizophrenic having an episode who either played with themselves or their partner.
T-shirt me baby!!
Or <Pigger> was on the scene, flying solo.
Boom! T-shirt me again Kappa
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Or <Pigger> was on the scene, flying solo.
FailFish
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You say:
"At the end of the party, Jason asked everyone in attendance"
Can we make any assumption that not everyone was still at the party when Jason asked?
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You say:
"At the end of the party, Jason asked everyone in attendance"
Can we make any assumption that not everyone was still at the party when Jason asked?
No. He asked everyone who attended. Nobody left before he asked.
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I tutored math, and still can't get this problem, but here goes an incorrect guess.
X is the variable for a typical person at the party who didn't know someone. Jason asked 9 different people
So for this I will say there were 9 people besides Jason at the party. So all the variables add up to that. Since everyone has a different answer it's that every single number from 1 to 9 was used. Because a person of the 9 can't use themselves we know it has to be at least X-1, and since every person has a different number we can go X-1 through -9.
So X-1+X-2+X-3+X-4+X-5+X-6+X-7+X-8+X-9=9
So 9X-45=9
So 9X=54
So 9X=6
The answer is 6. Anna played with 6 different players.
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I think I figured it out, but for real this time.
Anna played with 4 different people.
I'll get to the explaining after I eat dinner.
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I tutored math, and still can't get this problem, but here goes an incorrect guess.
X is the variable for a typical person at the party who didn't know someone. Jason asked 9 different people
So for this I will say there were 9 people besides Jason at the party. So all the variables add up to that. Since everyone has a different answer it's that every single number from 1 to 9 was used.
So X-1+X-2+X-3+X-4+X-5+X-6+X-7+X-8+X-9=9
So 9X-45=9
So 9X=54
So 9X=6
The answer is 6. Anna played with 6 different players.
"Math" may be too broad a term to rule out, but there's not really any arithmetic or algebra involved here.
Your reasoning is incorrect. I guess I should refrain from commenting on the numerical answers, otherwise people could eliminate all the wrong ones just by offering faulty reasoning for them.
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OK, one last guess.
The answer is unsolvable. Anna could have very well known every single person at the party, and therefore not played with a single person. Or Anna could not have known 8 people and therefore played with 8 people. So the answer is a number from 0 to 8. And that is my final guess, and reasoning.
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I think I figured it out, but for real this time.
Anna played with 4 different people.
I'll get to the explaining after I eat dinner.
As has already been established, Jason received 9 different answers, and those answers were 8,7,6,5,4,3,2,1, and 0.
For it to be possible for both 8 and 0 to have been given as answers, the person who played with 8 different people must have been married to the person who played with 0. i.e. No one played with themselves or their partner, so if someone played with 8 different people, they played with everyone except their partner, and thereby everyone except the partner of the person who played 8 must have played at least one game.
Thus 8 and 0 are paired, which means Anna is neither because Jason isn't answering.
Holding to the same logic, the person who played with 7 other people played with everyone except their partner and the person who played with no one, forcing all remaining party goers to have played with at least 2 different people. Therefore, the person who played 7 other people must be married to the person who played with only 1.
...6 and 2 are paired
...5 and 3 are paired
And 4 is left unpaired, so it must be Anna.
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Ross for the win
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Ross for the win
Not unless VON's given name is, by coincidence, Ross as well.
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I think I figured it out, but for real this time.
Anna played with 4 different people.
I'll get to the explaining after I eat dinner.
As has already been established, Jason received 9 different answers, and those answers were 8,7,6,5,4,3,2,1, and 0.
For it to be possible for both 8 and 0 to have been given as answers, the person who played with 8 different people must have been married to the person who played with 0. i.e. No one played with themselves or their partner, so if someone played with 8 different people, each of the people they played with played at least that one game.
Thus 8 and 0 are paired, which means Anna is neither, because Jason isn't answering.
Holding to the same logic, the person who played with 7 other people played with everyone except their partner and the person who played with no one, forcing all remaining party goers to have played with at least 2 different people. Therefore, the person who played 7 other people must be married to the person who played with only 1.
...6 and 2 are paired
...5 and 3 are paired
And 4 is left unpaired, so it must be Anna.
We have a winner! PM me your address and shirt size to claim your prize.
Reasoned slightly more thoroughly:
Because, obviously, no person played Donkey Kong with himself or herself, or with his or her spouse, nobody played Donkey Kong with more than eight other people. And since nine people played Donkey Kong with different numbers of people, these numbers must be 0, 1, 2, 3, 4, 5, 6, 7, and 8. We will refer to the people that gave those answers correspondingly as Person 0, Person 1, ... , and Person 8 respectively.
Person 8 played Donkey Kong with everyone except with his/her spouse. Therefore everyone at the party except for Person 8's spouse played Donkey Kong with at least one person. Therefore, the spouse of Person 8 must be Person 0.
Person 7 played Donkey Kong with everyone except his/her spouse and Person 0. Therefore everyone at the party except Person 0 and Person 7's spouse played Donkey Kong with at least 2 people (Person 7 and Person 8 ). Therefore, the spouse of Person 7 must be Person 1.
Person 6 played Donkey Kong with everyone except his/her spouse and Person 0 and Person 1. Therefore everyone at the party except Person 0, Person 1, and Person 6's spouse played Donkey Kong with at least 3 people (Person 6, Person 7, and Person 8 ). Therefore the spouse of Person 6 must be Person 2.
Person 5 played Donkey Kong with everyone except his/her spouse, Person 0, Person 1, and Person 2. Therefore everyone at the party except Person 0, Person 1, Person 2, and Person 5's spouse played Donkey Kong with at least 4 people. Therefore the spouse of Person 5 must be Person 3.
The only person left that answered Jason is Person 4, which must be Jason’s wife, Anna. The answer is: Anna played Donkey Kong with 4 people.
The chart below shows the scenario. The married couples are denoted by grey-shaded ovals. The colored lines connect the people who played Donkey Kong with each other.
(http://www.slither-gdi.net/chart.jpg)
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Ross for the win
Not unless VON's given name is, by coincidence, Ross as well.
Sorry, I was confused by homerwannabee's use of Ross' name in his sig. (And the absence of Ross' name in his own sig.)
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gg Ross
Once I fucked up I got cocky about what I "knew" and failed to pull out. BibleThump
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gg Ross
Once I fucked up I got cocky about what I "knew" and failed to pull out. BibleThump
And it's ironic how the point you got confused on is so closely related to the key that makes all the correct reasoning start to turn.
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A couple of followup challenge problems to make sure folks understand the original solution:
1) The first party Weehawk asked about above was held in the "dark ages" of Donkey Kong gaming, back before the King of Kong, the Kong Offs, and the recent explosion in the number and quality of Donkey Kong players and CAG sites like DKF. So it had fairly few players. After the Kong Off 4, Richie Knucklez hosted a week-long blow-out party like the first DK party, but this one had a total of 50 married couples, including once again Jason and Anna Cram. Also in attendance were George Strain and his stunningly beautiful wife Svetlana. The same rules applied as at the first party where each player played one game of doubles against anybody they didn't know, and no extra games were played. George played 5 more than half as many games as Svetlana. As the party wrapped up Jason once again inquired of all the other players the number of games they had played. Stunningly, he again got a different answer from everyone he asked!
- How many more games did Jason play than George?
2) John Lexmark's best friend Marcus Elswiner had an arcade party at his house. Marcus, Jason and Anna Cram, Dean Saglio and his wife, John Lexmark and Flossy the Sheep, and 3 other couples attended. As a surprise to all his guests, Marcus unveiled a new addition to his game room: a Zoo Keeper arcade game. Marcus ran a "mingler" party game of his own invention that was somewhat similar to the DK party games, except each person played one game of Zoo Keeper doubles with each person they had previously met, and no games with people they hadn't already met. No other games were played. Everybody enjoyed watching all the games, as a number of excellent players were in attendance and everybody tried their best.
Unfortunately, Marcus was murdered before he had played any games. At the end of the party and before the police showed up, John asked all the other still living players how many games they had played. He received various answers, the sum of which was less than 42 and the product of which was less than 16. Nobody played less games than Flossy the Sheep.
- How many games of doubles Zoo Keeper did John Lexmark play? Provide an explanation.
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2) John Lexmark's best friend Marcus Elswiner had an arcade party at his house. Marcus, Jason and Anna Cram, Dean Saglio and his wife, John Lexmark and Flossy the Sheep, and 3 other couples attended. As a surprise to all his guests, Marcus unveiled a new addition to his game room: a Zoo Keeper arcade game. Marcus ran a "mingler" party game of his own invention that was somewhat similar to the DK party games, except each person played one game of Zoo Keeper doubles with each person they had previously met, and no games with people they hadn't already met. No other games were played. Everybody enjoyed watching all the games, as a number of excellent players were in attendance and everybody tried their best.
Unfortunately, Marcus was murdered before he had played any games. At the end of the party and before the police showed up, John asked all the other still living players how many games they had played. He received various answers, the sum of which was less than 42 and the product of which was less than 16. Nobody played less games than Flossy the Sheep.
- How many games of doubles Zoo Keeper did John Lexmark play? Provide an explanation.
Zero. Lexmark doesn't play in public.
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1) The first party Weehawk asked about above was held in the "dark ages" of Donkey Kong gaming, back before the King of Kong, the Kong Offs, and the recent explosion in the number and quality of Donkey Kong players and CAG sites like DKF. So it had fairly few players. After the Kong Off 4, Richie Knucklez hosted a week-long blow-out party like the first DK party, but this one had a total of 50 married couples, including once again Jason and Anna Cram. Also in attendance were George Strain and his stunningly beautiful wife Svetlana. The same rules applied as at the first party where each player played one game of doubles against anybody they didn't know, and no extra games were played. George played 5 more than half as many games as Svetlana. As the party wrapped up Jason once again inquired of all the other players the number of games they had played. Stunningly, he again got a different answer from everyone he asked!
- How many more games did Jason play than George?
Finally, some algebra!
The pairings are 98 and 0, 97 and 1, 96 and 2, 95 and 3, down to 49 and 49.
Jason and Anna both played 49 games.
X = # of games played by George
Y = # of games played by Svetlana
Equation #1: X + Y = 98
Equation #2: Y / 2 = X - 5
Multiply equation #2 by 2: Y = 2X - 10
Plug Y back into equation #1: X + 2X - 10 = 98
Simplify and solve for X: 3X - 10 = 98, 3X = 108, X = 36
Jason played 13 games more than George (49 - 36).
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2) John Lexmark's best friend Marcus Elswiner had an arcade party at his house. Marcus, Jason and Anna Cram, Dean Saglio and his wife, John Lexmark and Flossy the Sheep, and 3 other couples attended. As a surprise to all his guests, Marcus unveiled a new addition to his game room: a Zoo Keeper arcade game. Marcus ran a "mingler" party game of his own invention that was somewhat similar to the DK party games, except each person played one game of Zoo Keeper doubles with each person they had previously met, and no games with people they hadn't already met. No other games were played. Everybody enjoyed watching all the games, as a number of excellent players were in attendance and everybody tried their best.
Unfortunately, Marcus was murdered before he had played any games. At the end of the party and before the police showed up, John asked all the other still living players how many games they had played. He received various answers, the sum of which was less than 42 and the product of which was less than 16. Nobody played less games than Flossy the Sheep.
- How many games of doubles Zoo Keeper did John Lexmark play? Provide an explanation.
As a matter of general principle, I don't participate in any Zoo Keeper threads.
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How many more games did Jason play than George?
Jason played 13 games more than George.
Correct. And well reasoned! The keys to getting started on this problem are:
1) Understanding that the exact same reasoning from the original problem can be generalized and applies regardless of the number of couples that played.
2) That reasoning dictates that each couple will have played a sum of games which is 2 less than the number of total players.
3) That reasoning also dictates that both Anna and Jason will have always each played (# of couples - 1) games.
After that it's just a bit of simple algebra.
-George
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How many games of doubles Zoo Keeper did John Lexmark play? Provide an explanation
Zero. Lexmark doesn't play in public.
Partial credit. The answer is correct and was the most obvious part of the puzzle. The bigger mystery is really about the murder and its singularity.
The sordid tale of the party itself is a bit more complex than at first glance.
Lexmark shows up at Marcus' party with his friend Flossy not knowing what is in store that night.
Lexmark is surprised at the other guests that have shown up, then when he sees Marcus has acquired a Zoo Keeper he starts to sweat. When Marcus starts with his mixer game to get the party going, Lexmark panics, knowing he'll have to finally show his play in public. He starts to calm down a bit and think things through. Fortunately for him, the only two attendees he's met before are Marcus and Flossy.
As a true competitor, Lexmark could never play a public game of Zoo Keeper in front of Dean and Jason and reveal his precious secrets. Marcus clearly had to go.
But what about Flossy you ask? Flossy wasn't murdered? Lexmark had to play Flossy! Ah, yes, the problem with Flossy. As Lexmark sharpened his knives, he reread the rules of the mixer game and noticed a glitch he could exploit! The rules only required people to play games, and Flossy was a sheep!
As the police showed up, Lexmark thought he'd get away with it too, by blaming it on those "troublemakers" in the CAG community that had been persecuting him. Unfortunately for him those meddling kids over at the Donkey Kong Forums sorted the mystery out and brought him to justice.
Disclaimer: All characters appearing in this work are fictitious. Any resemblance to real persons, living or dead, is purely coincidental.