Food comp 9/10ish
TV ClubHouse: Archive: Food comp 9/10ish
Prisonerno6 | Wednesday, September 10, 2003 - 02:25 pm     Sounds like it involves logic puzzles of some sort. They are talking about having to weight balls in groups and (I think) figure out which is the heaviest with only X number of weighings. It sounds like that is only one puzzle they need to solve. Erika said, "It sounds like we won't have food this week..." |
Justshirley | Wednesday, September 10, 2003 - 02:31 pm     Oh, fabulous! Just what we live feed watchers need -- for these few remaining HGs to be sluggish and sleep even more than they already do. |
Cas | Wednesday, September 10, 2003 - 02:32 pm     From Jokers website: Suppose you had 8 billiard balls, and one of them was slightly heavier, but the only way to tell was by putting it on a scale against another. What's the fewest number of times you'd have to use the scale to find the heavier ball? ....my guess is three...doesn´t seem that hard but maybe I´m missing something. |
Prisonerno6 | Wednesday, September 10, 2003 - 02:33 pm     BWAHAHAHAHA!!!! *giggle* *splort* *guffaw* Good one... Oh, sounds like it's just a competition for "the meal of your choice." They get food regardless of whether they solve the puzzle. |
Lurknomore | Wednesday, September 10, 2003 - 02:33 pm     They just said that John said they get real food this week...this comp is for them to get to choose what they want to eat tonight |
Rainwoman | Wednesday, September 10, 2003 - 02:34 pm     Jun keeps speaking in the "you guys" are going to have to do this or that for the food comp, while Erika is speaking in the "we're going to have to" do this or that. Erika is in for a rude awakening. |
Prisonerno6 | Wednesday, September 10, 2003 - 02:35 pm     It sounds like they only have two weighings in which to do this. |
Prisonerno6 | Wednesday, September 10, 2003 - 02:37 pm     Oh, I've got it in two. You weigh three balls against three balls, leaving two out. If the ones you weigh match in weight, you know the heavier ball is one of the two left. You weigh those two to find out which is heavier. If one group of three is heavier than another, you weigh two of those balls. If they match in weight, the heaviest ball is the one left out of that three. If they don't match, you have the heaviest ball weighed. Two weighings -- 3 against 3 and then 1 against 1. |
Cas | Wednesday, September 10, 2003 - 02:37 pm     Just went back and saw that...I don´t see how it´s possible in just two weighings?? Unless you can slowly put the balls on one at a time ;D |
Cas | Wednesday, September 10, 2003 - 02:40 pm     Clever! |
Prisonerno6 | Wednesday, September 10, 2003 - 02:41 pm     Now you know why I got an 800 on the logic portion of my GREs. |
Prisonerno6 | Wednesday, September 10, 2003 - 02:43 pm     Erika's on the right track... |
Dan | Wednesday, September 10, 2003 - 02:49 pm     Or leave 3 out, weight 2 against 2. If weight is different, weight the 2 balls from the heaviest stack to find the heaviest ball. If the first weighing is even, weight 2 of the 3 balls left out at first. If one of those 2 balls is heaviest, there you go. If weight is even the third ball is heaviest. |
Cas | Wednesday, September 10, 2003 - 02:51 pm     But there are eight balls, not seven. Prisonerno6, the GRE wizard, has it. |
Crossfire | Wednesday, September 10, 2003 - 02:54 pm     If you leave 3 out, and go 2 vs 2. You will have to hope no one catches you firing the 8th ball over the fence. Additionally, if that 8th ball laying in the parking lot it the heavy one, you are in a pile of trouble.  |
Fan3 | Wednesday, September 10, 2003 - 02:54 pm     okay my question if you know you are leaving why help the remaining people get good food? she (Erica) has done told Rob she is sure she is leaving, Boy if it was me I would be trying to mess with their minds and do it backwards or anything else to screw them up. |
Cliotheleo | Wednesday, September 10, 2003 - 02:56 pm     Because she's a nice person, same reason she cleaned the house this morning. I wouldn't have done it, but I am both mean and lazy, two qualities Erika does not possess. |
Cas | Wednesday, September 10, 2003 - 02:56 pm     I know, I´d let them suffer but I´m bitter. |
Torch | Wednesday, September 10, 2003 - 03:10 pm     This is an old math problem and Prisonerno6 is correct. You can actually find the heavest/lightest of 9 balls in two weighings. |
Spear | Wednesday, September 10, 2003 - 03:27 pm     The version I've heard before is 12 balls in three weighings, but with the additional constraint that you don't know whether the odd ball is light or heavy (or if there is even an odd ball). |
Prisonerno6 | Wednesday, September 10, 2003 - 03:48 pm     I would have figured it out on my own, told them I had the answer, and then said I wouldn't share it unless they kept me in the house. Then let them stew wondering if I was telling the truth. And if they didn't keep me, tell them the answer on the way out to prove I knew it, and tell them I hope they enjoy all the octopus and gefilte fish this week. BTW, on the drive home from work, I realized this worked for 9 balls as well, as Torch pointed out. However, I like the 8 balls instead, because your first instinct is to split them into two even groups. With 9, the leap to realizing you leave some balls out of the first weighing is easier to make. |
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