Pennies...really?? Spoiler 1) Weigh 17 by 17. (16 on side) 2a) If one is heavier then split that by 6 and 6 and weigh it (5 on side) or 2b) If not split the 16 on the side into 5 by 5 and weigh it (6 on side) 3aa) If one is heavier split that by 3 and 3 3ab) If not, split the by 2 and 2 3ba) If one is heavier split by 2 and 2 3bb) If not split by 3 and 3 4a) For 2, weigh the remainder 4b) For 3 left, weight by one and one. If both are balanced, the one unweighed is the counterfeit.
Didn't get it, this is close as I could come. Spoiler Weighing number one has 25 pennies on one sale, 25 one the other. Remove the lighter 25 load next to ground-side A of the scale, remove one penny off of the heavier load and place it next to ground-side B of the scale. Now weighing number two has loads of 12 pennies a piece against each other. If they're even, the one penny on ground-side B is the heavier one. Remove the lighter 12 load and dump it on ground-side A. Remove one penny from the heavy load and place it on ground-side C. Weighing number three has two loads of 6 pennies against each other. If they're even, the one penny on ground-side C is the heavier one. Remove the lighter 6 load and dump it on ground-side A. Take one penny off of the heavier 6 load and place it on ground-side D. Now you have 5 pennies from the heavier load. Take 1 penny-K from the 5-heavy scale X and place it on scale Y. Take one penny from 4-heavy scale X and place it on ground-side E. Take the 2 pennies from ground-side B and C and place them on 1 scale Y. Weighing #4, if scale X is heavier, penny K is the counterfeit.
THAT'S WHAT SHE SAID. ----------------------- About the pennies, my solution: Spoiler Divide into two piles: 20 in one pile, 30 in the other. Weighing 1, if you start with the 30: 10 vs. 10 10 in the hand then Weighing 2: 4 vs. 4 2 in the hand Weighing 3: 2 v 2 Weighing 4: 1 v 1 Done. -Starting with 20? ------------------ Weighing 1, if you start with the 20: 6 vs. 6 (8 in the hand) then Weighing 2: 2 vs. 2 two in the hand Weighing 3: 1 v 1 (one in the hand) DAMN YOU, heypartner, couldn't you just have pointed to the one in my post in the puzzles thread? About the numbers, I think it has to do something with Spoiler highest two-worded number where letters don't repeat? Is that right, heyp?
Spoiler Beat me to it. Though I think it's just numbers with no repeating letters in general. If you had more words, you'd have repeated letters. Nice puzzle! My puzzle is: If you have a weighted coin where the chance of flipping heads is "p" and tails is "1-p", but you don't know the value of "p", how can you still use the coin to simulate a 50/50 chance?
Professor in my Business Law class (many years ago) gave this puzzle on first day of class. Out of couple hundred student I was only one find solution or possible the only one who tried to solve it. "The Puzzle: You have 12 balls identical in size and appearance but 1 is an odd weight (could be either light or heavy). You have a set scales (balance) which will give 3 possible readings: Left = Right, Left > Right or Left < Right (ie Left and Right have equal weight, Left is Heavier, or Left is Lighter). You have only 3 chances to weigh the balls in any combination using the scales. Determine which ball is the odd one and if it's heavier or lighter than the rest. How do you do it?" Website with solution Spoiler http://www.mathsisfun.com/pool_balls_solution.html
^^^ ??? Dude. It's the same puzzle and I don't need the website. 1st weighing: 4 on the left, 4 on the right, 4 in your hand. 2nd weighing: 2 in the scale, 2 in your hand. 3rd weighing: 1 on each scale.
Spoiler You flip the coin twice, but only consider two states valid. Head Tails and Tails Head. The combinations Heads Heads and Tails Tails is invalid. The probability of getting head tails and tails head are the same P-P^2 so you have a 50 50 shot of getting one or the other.
The scenario I placed was for the outcome that is the one you can use as a basis for the others either way it goes. You're welcome.
Your first weighting is correct. Reminder you have identify if ball is heavier or lighter and tell which ball it is. Start with 1st scenario, when scale tips. You now 4 ball that can be heavier, 4 ball that can lighter and 4 balls that you know is equal weight. With your method explain how can you identified which ball is odd one and if heavier or lighter?
Don't you understand? You just go either way, no matter what, and you would have FOUR in which one is lighter, FOUR in which one is heavier, and FOUR in which you don't know because it's not on the scale. From there, you just GO. Take it... and go. TAKE IT and GO. It's logic from there. 4 on the scale left. 4 on the scale right. 4 on your hand. 1st weighting: Scale tips (take your pick) left. Use those four on the left for the next (2nd weighing). Now put two of those on the left, two on the right. Scale tips right, now use those two to determine your last weighing (3rd). You're done. The only thing you need to worry about is to have enough in your hand and NOT on the scale at the beginning to weed out a large group that you can't figure out, and that's FOUR at the maximum. You get it now?
There are multiple solutions if your starting point is 50. Other numbers won't have multiple solutions and a specific algorithm is needed to find the minimal answer. For example, 68 coins instead of 50.
1) 4 heavy, 4 light, 4 normal 2) 2 heavy vs 2 heavy = scale doesn't tip they equal 3) Your last weighting you have 4 light one left which one is the light one?
You still don't get it. You don't have to place all of them on the scales! Check it: So far, so good ^ but, remember, 8 are on the scale, 4 are in your hand. That's where you go wrong, man. You don't do "2 vs. 2", you do two on the scale (one on each side), and two in your hand. If they're the same, then the heaviest one is in your hand. In the 3rd weighing, do those two. DONE. Got it?
Two railroad trains (on different tracks) are approaching one another on a straight railroad line, each train traveling at a speed of 30 mph. Suppose a fly -- a very fast-flying fly -- traveling at 60 mph, starts from the front of train A when the two trains are ten miles apart. It flies until it reaches the front of train B, turns around immediately and flies back to train A, where it turns around again and starts back to B. The fly continues to travel back and forth between the two trains, making shorter and shorter journeys, until the trains pass one another on the tracks. What is the total distance flown by the fly?
Spoiler 10 miles. It will take 10 minutes for the trains to pass eachother, so if the fly is traveling at 60 mph during those 10 minutes it will have flown a distance of 10 miles.
