.Nu-13 Posted December 25, 2010 Report Share Posted December 25, 2010 Yeah, my sudden obsession about 999 for DS popped out :P [spoiler=Rules]Any number of people can join. However, from among them, I'll choose the 9 that will participate.Each of participants will receive their number, be "locked" and have to "get out". This means that each of you will receive individual task. If you don't complete it, you're eliminated. This will be elimination round.There will be 9 "numbered doors", the 9th being the final one. During each round, participants, among themselves, will decide which door they want to take, except 9th. However, there's a condition.Only 3-5 people can "enter" the door. But, the problem is, that not everyone can enter one door. They "digital root" of entering people must be equal to the number on door. What is the "digital root"? It's the sum of numbers of entering people. If after adding them you get 2-digit number, add both digits. Then, keep doing this until you get the 1-digit number. This is the "digital root". In each door, there will be different task.However, you have to think the teams entering so that as many as possible people will "survive". If it's not possible to get needed digital roots from all people, you have to "sacrifice" some people the way that other people can enter certain door.There are even more complications. If at some point, there will be no way that people can enter a door, it means that all of you are eliminated and loose. No one gets prizes. So, this is a game of risk and intelligence. [spoiler=Participants]Awesome Ryu~Shadow of Alpha~Cyber-Valley™ newfie sharkSpoonMonkeymadmanencapturerMasterCyclone! [spoiler=Prizes]All people that come through the 9th door will get 1000 points and 2 reps. [spoiler=Entry Fee]No fee [spoiler=Doors]1st2nd3rd4th5th6th7th8th9th If there are still any questions, feel free to ask. Link to comment Share on other sites More sharing options...
Tee-Jaie Posted December 25, 2010 Report Share Posted December 25, 2010 Im interested in joining, but, im just a little confused. Explain, just a wee bit slower. Link to comment Share on other sites More sharing options...
.Nu-13 Posted December 25, 2010 Author Report Share Posted December 25, 2010 Ok, I'll give an example All 9 people passed eliminations, so we have1 2 3 4 5 6 7 8 9 They decided to split into 3 3-people groups 1st group - 1 3 52nd group - 2 4 63rd group - 7 8 9 1 + 3 + 5 = 82 + 4 + 6 = 12, 2 + 1 = 37 + 8 + 9 = 42, 4 + 2 = 6 So, 1st group can enter door 8, 2nd can 3, and 3rd can 6 Let's assume that the 1st group didn't make it. The following people are left:2 4 6 7 8 9 In that case, they must split into 2 3-people groups. Assuming that all 6 made it to the door 9, and only 5 people can enter, they must choose 1 person to be eliminated Link to comment Share on other sites More sharing options...
Tee-Jaie Posted December 25, 2010 Report Share Posted December 25, 2010 Oh, I get it, I'm in B) Link to comment Share on other sites More sharing options...
.Nu-13 Posted December 25, 2010 Author Report Share Posted December 25, 2010 If so, accepted :) Link to comment Share on other sites More sharing options...
~Shadow of Alpha~ Posted December 25, 2010 Report Share Posted December 25, 2010 Alright, I'm in. Link to comment Share on other sites More sharing options...
.Nu-13 Posted December 25, 2010 Author Report Share Posted December 25, 2010 Sure ^^ At least 7 more people... Link to comment Share on other sites More sharing options...
Cyber-Valley™ Posted December 25, 2010 Report Share Posted December 25, 2010 im in sounds great Link to comment Share on other sites More sharing options...
~Shadow of Alpha~ Posted December 25, 2010 Report Share Posted December 25, 2010 Great. Merry Christmas Link to comment Share on other sites More sharing options...
Newfie Kuriboh Shark Posted December 25, 2010 Report Share Posted December 25, 2010 i shall join. its about time somebody posted this. Link to comment Share on other sites More sharing options...
.Nu-13 Posted December 25, 2010 Author Report Share Posted December 25, 2010 "about time"? And both accepted Link to comment Share on other sites More sharing options...
Spoon Posted December 25, 2010 Report Share Posted December 25, 2010 Wow, this sounds horribly complicated... I'll join :P Because it's also an interesting concept. One question though: If we're going in as teams, we also have to solve the "problems" in teams, meaning we create the card together, right? I've made some bad experiences with this... but it's worth a try nontheless :) Link to comment Share on other sites More sharing options...
.Nu-13 Posted December 25, 2010 Author Report Share Posted December 25, 2010 Yes, you work on cards together, unless the challenge specifies that you have to work separetaly Link to comment Share on other sites More sharing options...
Newfie Kuriboh Shark Posted December 25, 2010 Report Share Posted December 25, 2010 just focus on the card requirements, thats what matters ;) oops, did i just give away a secret? Link to comment Share on other sites More sharing options...
.Nu-13 Posted December 25, 2010 Author Report Share Posted December 25, 2010 Ok, I have to go for today... guess what I'll be playing ^^ Link to comment Share on other sites More sharing options...
Newfie Kuriboh Shark Posted December 25, 2010 Report Share Posted December 25, 2010 depends. what did you get for christmas? Link to comment Share on other sites More sharing options...
.Nu-13 Posted December 25, 2010 Author Report Share Posted December 25, 2010 I thought that contest based on game and including it's title in first post will be enough to guess :/ Link to comment Share on other sites More sharing options...
Monkeymadman Posted December 25, 2010 Report Share Posted December 25, 2010 I'll join, but i might get confused and you'll have to explain it to me again. I guess i'l figure it out eventually. Link to comment Share on other sites More sharing options...
encapturer Posted December 26, 2010 Report Share Posted December 26, 2010 999? Interesting. I'll join. Link to comment Share on other sites More sharing options...
.Nu-13 Posted December 26, 2010 Author Report Share Posted December 26, 2010 Both accepted Link to comment Share on other sites More sharing options...
.Nu-13 Posted December 27, 2010 Author Report Share Posted December 27, 2010 BUMP Damn, I frogot 3 things... 1. Door 9 are just like other ones. They still have 3-5 people limit, and the entering peoples' digital root must be equal 9.2. You can't choose doors at will. You will be given a set of doors each time.3. Your numbers won't be told you directly. They'll be coded Link to comment Share on other sites More sharing options...
The Sleeping Dude Posted December 27, 2010 Report Share Posted December 27, 2010 joining!! Link to comment Share on other sites More sharing options...
.Nu-13 Posted December 27, 2010 Author Report Share Posted December 27, 2010 YES! Accepted 1 MORE TO START! Link to comment Share on other sites More sharing options...
Brinolovania Posted December 27, 2010 Report Share Posted December 27, 2010 What the heck. COUNT ME IN TC. Link to comment Share on other sites More sharing options...
.Nu-13 Posted December 27, 2010 Author Report Share Posted December 27, 2010 YES! 9 people!Ok, gimme a moment to assign code numbers... Ok, here are your code names/numbers. Your name relates to number somehow. If you figure out your number, PM me it. Awesome Ryu - Snake~Shadow of Alpha~ - CloverCyber-Valley™ - Acenewfie shark - PlutoSpoon - JuneMonkeymadman - Luckyencapturer - SantaMasterCyclone - Star! - Lotus Link to comment Share on other sites More sharing options...
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