OK, now I see Baron is online, so it's time to bring out the heavy artillery: You have 12 pennies. 11 of them weigh exactly the same. 1 of the 12 is either heavier or lighter than the other 11. You don't know if it is heavier or lighter. Just that it is of a different weight than the others. You have a balance scale. You can use the balance scale only 3 times. How do you go about finding the one unequally weighted penny using just the three weighings? (disclaimer: I cannot figure this out so far, and I know there are 2 solutions). Go Carl! RS7
I'm a slow one - I needed a pad of paper and it took me more than 10 minutes. Don't think I could have thought of it in a car going to play golf in under 5 minutes. rs7, you must be brilliant. Sorry Publias - didn't mean to offend - your answer was good enough. I just wanted to see rs7's next problem - the one he hasn't solved yet. Where is it? Carl
I am not brilliant. I figured it out by being simple minded. I couldn't do the higher math, so I just made the bridge an 8 hour walk. That way if I knew I had 5 hours to go the long way, and 3 hours the short way. 40 miles, and 24 miles. The 2 hour difference in my walk meant the train had to cover 64 miles in 2 hours. 32 MPH. How did you figure it out? Rs7
I used algebra: (3/64)x=y/t (5/64)x=(x+y)/t with these two equations you find that (2/3)y=x plugging back in to the first equation, t = 32 in the equations above, x = distance of entire trestle, y = distance of train to beginning of trestle, and t = miles per hour of train (the 8mph of the walker is built into the equations). Carl
So that proves it...you are definitely smarter than me. I used logic, not math. I don't even understand the formula OK, now on to the big 12 penny problem. Lots of competition now. I see that mrSub, Darkhorse, Magna, Babak and Fasterpussycat have signed on since I last looked. But Baron left....so maybe he is consulting with his liquid hydrogen cooled supercomputer, which can't be used while his PC is turned on (a power consumption issue I would imagine). Waiting very patiently for an answer. Good luck to all. I will buy an Elite Trader T-shirt for whoever gets it first. (I would expect a deep discount if it turns out to be Baron). RS7
When first presented with this problem, I solved it after 20 minutes. There is only one solution as far as I know.
Well, I can do it in 3 tries 50% of the time, and 4 tries the other 50%. If I think a bit harder, then maybe I can get that 4 down to 3. But here is where I am at right now. First, divide the 12 pennies into three groups of four pennies each. Pick two of the goups and put them on the scale (one on each side of course). If the scale shows both groups weigh the same, then you can discard all eight pennies from both groups. That leaves four pennies. Weigh two pennies and record the findings. Then keep one of the pennies you just weighed on the scale, but switch the other penny for one that you haven't weighed. That will give you the relative weights of three pennies. If all three were equal, then the fourth is the different one. If the first two were equal, but the third was different, well then you know it was the third. Etc. The problem is that if the weighing of the first eight pennies shows that the different one is in those eight, then you only get to throw out the four that weren't weighed. Reducing the remaining eight down to the four I use in the above final two weighings would take me one extra weighing using my current logic. Oh well, its getting late.