How far apart are the tracks from each other? EDIT: What I mean, is... Spoiler :grin: Sorry for the profanity, but I couldn't think of any other way to convey my question.
This is a clever 'misdirection' puzzle, I like it. Spoiler The only important factors here are the speed of the fly and how long the fly will be flying at that speed. None of the other factors matter at all. Since we are already given the speed of the fly (60mph), all we need to know then is how long the fly will be flying. The amount of time is the amount needed for the two trains to meet. Two trains approaching each other at 30mph is the same thing as one train traversing the entire distance at twice the speed - ie 60mph.. amusingly enough, the exact speed of the fly as well. To travel 10 miles at 60mph will require 12 minutes (1/6th of an hour). Since we already know the fly will be traveling at 60mph for the entire 12 minute duration, we can stop here - the fly will travel the full ten miles as the riddle is laid out. The length of each segment the fly travels before reversing is immaterial. So there you go. Cool puzzle, I like it, had not seen that one before.
Spoiler Your answer of 10 miles is correct. However, it takes 10 minutes to travel 10 miles at 60 mph, not 12 minutes.
Spoiler What was I smoking? In my own crazy world, it takes 12 minutes, that's all I can say.. :grin:
So your 2nd weighing is? 1 v 1 What about 4 light ball? You 6 ball in hand no 2. You 2nd weighting should be 3 vs 3. Why don't you just look at answer?
F***. I don't understand why you're still asking... but... it seems I'm getting through to you, man, so I will oblige, since you're asking nicely: YES. And two in your hand. In this second weighing, one of those two on the scales on one side will be heavier, or both the same. If the former, then for the last and third weighing, you will have not use the ones in your hand, and you're done. If the latter, then for the last and third weighing, you will have one versus one, and that will be easy... (or should I tell you what happens then?) Mother f#$&)(... man... by your lack of verbs, I can tell you don't speak English well, so I can see how you probably don't understand. Sorry this is so confusing. FOR THE LAST TIME, man... you don't need to have all on the scales. You can have some off the scales!!
In preparation for entering the Santiago Derby you have to take your 25 llamas and determine the fastest 3 llamas in order. You don't have any timekeeping device so you can only judge their speeds relative to each other. You have a race track that only can race 5 llamas at a time. What is the fewest number of races you need to hold to determine the top 3?
Your method wouldn't work for that problem, only because you don't know if the odd ball is heavier or light than the rest. If the scale tips with that first four on each side, and you just automatically pick the heavier side to find that odd ball, you wont find it on that side if the odd ball is actually light(it would be one of the four on the lighter side). Likewise, if you just pick the lighter side, you might be looking in the wrong place again because the odd ball might be heavier and might be on the heavy side.
Spoiler Best I could do was 9 races. 1st round of 5 races to get all of the llamas against each other and eliminate 10 of them. Then one race for all the 3rds, eliminate the bottom 4. One race for all the 2nds, eliminate the bottom 3. One race for all the 1sts, eliminate the bottom two, and remove the top one because he is the "winner, then race the one 3rd, two 2nds, and two 1sts to get the last two spots.
Spoiler 7 races is the lowest i could come up with. Race 5 group of 5 llamas. This eliminates 10 llamas (the last 2 in each race). Then race the 5 winners (6th race). This gives you the fastest llama. We had 15 llamas and since we don't need to race the fastest llama we are down to 14 llamas. We can also eliminate the 2 slowest from the llamas from our champions race and the llamas they beat in their preliminary race. This would bring us down to 8 llamas. We can also eliminate the llamas that lost in the preliminary round to the third place finisher of the champion round (the best they could be is 4th and 5th). This brings us to 6 llamas left. Finally we take the second place llama from the champions race and eliminate the llama that came in third behind him in the preliminary round (the best he could be is 4th). This brings us to 5 llamas. Race (7th race) the five and take the best two and they are second and third.
Llama problem: Spoiler The lowest I could come up with is 12. Your first 5 races (5x5=25) determine the fastest 15 llamas, because in each of those races, the last two are eliminated overall. Your next 3 races (5x3=15) determine the fastest 9 llamas, because the last two in each of those races are eliminated overall. You can only do 2 races next to only fit 9 llamas, so you get 6 left. Next, you do 1 race of 5 llamas, leave one out. Remember the order of these. Next you do 1 race of the third place above vs the one left over, and if it beats that third place, that one is the 3rd fastest, not the previous 3rd place, otherwise the last three are the winners. Yeah, no, I don't think your assessment of my answer is right. I believe I'm still right with these methods. I'd like to see, however, where the original question was raised, to see if he either wrote it correctly or just added the "lighter or heavier". I think that's where I might be answering one way and you both think I'm wrong. EDIT: Else, look at the solution homeboy gave... it's just like my answer.
If one chicken says, "All chickens are liars" is the chicken telling the truth? Spoiler Chicken can't talk
The lighter or heavier part of the oddball is what makes it complex. If not for that, your solution would work 100%. Look a little closer. Your solution works in the case where the 8 balls balance out at the beginning. To a tee, exactly as you described. However, look at the 2nd and third situations. Those are the cases where the balance isn't even. In those a lot more is going on. Not only do they throw in some of the balls they know are "normal", but there is a lot of ball switching going on to determine a) which is the odd ball b) is it heavy or light I actually solved that one with pictures rather than words so I can see where its easy to gloss over that text as it's a little involved.
