Do you have a logic/puzzle to post? I love these My entry Today on Car Talk,,,you know they have a question of the week which is usually a car question,,,well they are geeks like us so every once in a while they ask a logic question Today it was You have 50 pennies but one is counterfeit and weighs more than the real pennies. You have to find the counterfeit Penny. You have a balance with two trays as normal. And you are allowed four weighings of the 50 coins to find the counterfeit heavy one Can you do it? They will give an answer next week. Please spoiler answers and provide more puzzles to solve. If you heard the show my hint is to solve the base case first
I need 5 weighings: Spoiler Step 1: split stack of coins into 25 and 25 and weigh them against each other. One stack will weigh more. Step 2: Split the heavy stack in 12, 12 and 1. Weigh the two 12 stacks against each other. If they are equal then the single penny is the heavy one. If they are not equal, one 12 stack will weigh more. Step 3: Split the heavy 12 stack into 6 and 6 and weigh them against each other. One 6 stack will weigh more than the other. Step 4: Split the 6 stack in two stacks of 3 and weigh them against each other. One will weigh more. Step 5: From the heavy 3 stack, separate into 1, 1, 1 pennies. Weigh two of the pennies against each other. If they weigh the same, the single un-weighed penny is the heavy one. Else, the balance will show the single penny that weighs the most. Are you sure it was 4 weigh ins? I'll think about it a bit more. Edit: We had fun in this thread a few months back: http://bbs.clutchfans.net/showthread.php?t=232267
Spoiler 1. take a chance and take 32 pennies and throw the rest away because you're rich like dat. 2. Split what you have into 16 coins and weigh them. (first weighing) 3. Split the heaver side and split to 8 and weigh. (second weighing) 3. Split the heavier side to 4 and weigh. (third weighing) 4. Split the heavier side to 2 and weigh. (last weighing) Then you can weigh the last 2 with your hands. Spoiler I'm bad at math
Let's see if my explanation makes sense... is this right? Spoiler Start with 50 pennies... remove 8, now left with 42. First weighing: 21 vs. 21... if they are balanced, then one of the 8 is the counterfeit, use next 3 weighings to find fake. If they are unbalanced, then take the heavier one, remove other 21 (+ the first 8 since we now know they are real). Now left with 21 pennies. Remove 5 pennies, now left with 16. Second weighing: 8 vs. 8... if they are balanced, then one of the 5 removed is the counterfeit, use next 2 weighings to determine fake. If they are unbalanced, then take the heavier one, remove the other 8 (+ the first 5 since we now know they are real). Now left with 8 pennies. Remove 3 more pennies, now left with 5. Third weighing: 2 vs. 2... if they are balanced, then one of the removed 3 is the counterfeit. If this is the case, use the fourth weighing to figure out which of the 3 is counterfeit. If they are unbalanced, then take the heavier one, remove the other 2 (+ the first 3 since we now know they are real). Now left with 2 pennies. Fourth weighing: 1 vs. 1. Is this right??? I ran thru this pretty fast and didn't really double check the math lol.
Not sure if I solved it in the way it supposed to be solved, but think I kind of got it... Spoiler 1st weighing in blue 2nd weighing in red 3rd weighing in green 4th weighing in orange 1. Split stack into 20-20-10. 2. Weigh the 20 pennies against the other 20. a. If equal, split the stack of 10 by half and weigh 5 pennies against 5. b. Split the heavier 5 stack into 2-2-1 c. Weigh 2 pennies against 2 d. If equal, the remaining 1 penny is the fake. If not equal simply weigh the 2 pennies in the heavier stack against each other to find the heavier one 3. If the two 20 pennies stack are unequal in weight, split the heavier stack into 7-7-6 and weigh the 7pennies against the other 7. a. If equal, split the 6 stack into 3-3 and weigh them. b. Split the heaver stack into 1-1-1. Weigh any two of the 3 against each other. If equal, the odd one out is the fake. If not, the heavier coin is the fake. 4. If the two 7 stacks are unequal, a. Split the heavier stack into 3-3-1 and weigh 3 against 3. b. If the stack is equal, the odd one out is the fake. c. If stack is unequal, split the heavier stack into 1-1-1 and follow the steps in 3b.
Mingbling and el gnomo both got it. I might have to check the math because there are many ways to solve it at 50 pennies but you got the concept Now can you tell me the maximum pennies you can solve in 4 weighings Here's they math. Don't look at this if you want to try to figure out the maximum pennies you can do in 4 weighings Spoiler Discrete algebra tells us to solve the base case first. The base case is how many pennies can you solve in one weighing if the heavy one is present. The answer is three pennies. Two on the scale and one off the scale So what about two weighings? What is the most pennies. The answer is nine. Three on each side of the scale and three off That is 3 to the second power. So what about 3 weighings. Put 9 on each scale and 9 off. That is 27 3 to the third power So four weighings can solve 81 pennies. 3 to the 4th power Car talk used 50 instead of 81 to confuse us My solution was, well if I can solve 27 in three moves the let me move 26 pennies off the scale and so 1st weighing is 12 pennies against 12
Spoiler The max is 81. Spoiler Weigh 27 against 27. Weigh 9 against 9. Weigh 3 against 3. Weigh 1 against 1. It finally struck me to think of how many i could eliminate at the base case to distinguish 1 penny. The key to this problem is (# of balance pts + 1)^4.
Easy. Spoiler 1st measurement: 21 pennies on each tray. The next measurements must then be 4:4, 2:2 and 1:1. It all depends on if the first measurement is balanced or unbalanced. For example, if it were to balance then you know one of the remaining 8 pennies is counterfeit; if it is unbalanced, then you know 1 of the 21 pennies is counterfeit. Proceed along that logic.
srnm. Boo. Spoiler that So don't y'all enter your answers to car talk. I solved that in 5 minutes while driving on the freeway in my head. They had a much tougher one about 8 years ago about match sticks. That one took me two days. I'm on my cell phone now. Will post this later to make sure I remember And then another by Will Short the puzzle master too Please contribute your own So next question
This is from Will Short the puzzle master. If you don't know him he is the New York Times crossword puzzle editor These three numbers share a common characteristic...what is it? Five thousand is the largest number with this characteristic One sixth is the smallest positive number with it, and minus forty is the smallest number What is the characteristic they each share Hint. Will short asked this question not car talk
Remember to add your own questions and spoiler big hints and answers I'll post my favorite car talk logic question tomorrow
This is a like a standard compsci algorithms questions. Worst case it should take 4 tries. Spoiler You split them into three bins: 17,17,16 - Put the two 17s on the scales if they are even then it is in the 16 bin else it is one in the heavier bin. Do the same with 16 or 17 5,5,6 or 6,6,5. then 2,2,1 or 2,2,2 Worst case it takes 4 tries
Spoiler There are 3 results to a scale (left is heavier, right is heavier, and even). Therefore the algorithm would be to break the unknown into thirds and weigh two of these groups. For 50, if divided into 3rds we get, 17, 17, and 16. Put the two 17s on the scale. If one side is heavier then the coin is in that group. If they are equal, then the coin is in the 16 group. Therefore you are left with either 16 or 17 coins. IF you repeat this process again you will be left with either 5 or 6 coins (16 can be divided into 5,5,and 6. 17 can be divided into 5,6,and 6). IF you repeat it again, you will be left with 1 or 2. If you repeat for a 4th time, you get 1. Swoly asked the same question in the Riddles thread.
Spoiler (1) 50 => weigh( 16, 16 ), 18 Pile of 16 or 18 coins identified as having counterfeit. (2) 16 => weigh( 5, 5 ), 6 (2) 18 => weigh( 6, 6 ), 6 Pile of 5 or 6 coins identified as having counterfeit. (3) 5 => weigh( 2, 2 ), 3 (3) 6 => weigh( 2, 2 ), 2 Pile of 2 or 3 coins identified as having counterfeit. (4) 2 => weigh( 1, 1 ) (4) 3 => weigh( 1, 1 ), 1 On the 4th weighing, we can isolate which coin is the counterfeit, since we're left with a pile of 2 coins or 3 coins.
Haven't looked at the others, but here's my solution. Spoiler Take 6 pennies and set them aside. Split the remaining 44 pennies into 2 piles of 22 and weigh them. WEIGHING #1. If they are balanced, the counterfeit penny must be one of the 6 pennies you didn't weigh. Split those 6 into two piles of 3 and weigh them. WEIGHING #2-A. From the pile that weighs more, set one penny aside and weigh the other two. WEIGHING #3-A. If one weighs more than the other, that's your fake. If they're balanced, the penny you set aside is the fake. If one pile of the 22 pennies weighs more, then put 4 pennies aside, split the remaining 18 pennies into 2 piles of 9, and weigh them. WEIGHING #2 If they are balanced, the counterfeit penny must be one of the 4 pennies you didn't weigh. Split those pennies into two piles of 2 and weigh them. WEIGHING #2-B. Take whichever pile weighs more and weigh those two pennies against each other. WEIGHING #3-B. Whichever weighs more is the fake. If one pile of the 9 pennies weighs more, then put 3 pennies aside, split the remaining 6 pennies into 2 piles of 3, and weigh them WEIGHING #3. If one pile weighs more than the other, the counterfeit is in that pile. If the piles are balanced, the counterfeit is in the pile of 3 you set aside. Take the pile containing the counterfeit and set one penny aside and weigh the remaining 2 against each other. WEIGHING #4. If one penny weighs more than the other, that's the fake. If they are balanced, the penny you set aside is the fake. There you go.
