Author Topic: Arcade House Party Conundrum  (Read 13939 times)

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Offline homerwannabee

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Re: Arcade House Party Conundrum
« Reply #15 on: August 31, 2014, 05:55:21 pm »
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.
« Last Edit: August 31, 2014, 06:00:19 pm by homerwannabee »
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Offline VON

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Re: Arcade House Party Conundrum
« Reply #16 on: August 31, 2014, 06:01:41 pm »
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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Re: Arcade House Party Conundrum
« Reply #17 on: August 31, 2014, 06:02:08 pm »
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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Offline homerwannabee

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Re: Arcade House Party Conundrum
« Reply #18 on: August 31, 2014, 06:24:13 pm »
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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Offline VON

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Re: Arcade House Party Conundrum
« Reply #19 on: August 31, 2014, 06:28:39 pm »
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.
« Last Edit: September 01, 2014, 08:47:32 pm by VON »
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Offline Barra

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Re: Arcade House Party Conundrum
« Reply #20 on: August 31, 2014, 06:30:19 pm »
Ross for the win
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Online Weehawk

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Re: Arcade House Party Conundrum
« Reply #21 on: August 31, 2014, 06:32:55 pm »
Ross for the win

Not unless VON's given name is, by coincidence, Ross as well.
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Re: Arcade House Party Conundrum
« Reply #22 on: August 31, 2014, 06:42:43 pm »
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.

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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.

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Online Weehawk

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Re: Arcade House Party Conundrum
« Reply #23 on: August 31, 2014, 06:57:26 pm »
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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Offline ChrisP

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Re: Arcade House Party Conundrum
« Reply #24 on: August 31, 2014, 07:21:49 pm »
gg Ross

Once I fucked up I got cocky about what I "knew" and failed to pull out. BibleThump

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Online Weehawk

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Re: Arcade House Party Conundrum
« Reply #25 on: August 31, 2014, 07:26:00 pm »
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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Offline gstrain

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Re: Arcade House Party Conundrum
« Reply #26 on: September 01, 2014, 07:37:00 pm »
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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Offline VON

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Re: Arcade House Party Conundrum
« Reply #27 on: September 01, 2014, 08:32:04 pm »
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.

Offline stella_blue

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Re: Arcade House Party Conundrum
« Reply #28 on: September 01, 2014, 08:44:22 pm »
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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Offline stella_blue

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Re: Arcade House Party Conundrum
« Reply #29 on: September 01, 2014, 08:58:22 pm »
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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