Radical Edward Posted February 18, 2004 Posted February 18, 2004 I just thought that I would post some puzzles. These are pretty easy ones, but they are nice to get the brain cells ticking for a few moments. If anyone else has any Puzzles, please post them in this thread, and then I will add them to this Opening Post. If the thread is popular enough, I will split it later into a "Puzzles" and a "solutions to puzzles" thread to make for easier reading. (1) You have in front of you a balance scale and 18 apparently indentical coins. One of these coins is a forgery although it looks exactly like the other 17 coins: the only difference is that the fake coin weighs less than the normal coins. Your task is to sift out the coin in the least number of turns. (2) You have in front of you a balance scale and 12 apparently indentical coins. One of these coins is a forgery although it looks exactly like the other 11 coins: Unlike the previous puzzle, the coin may be heavier or lighter. Your task is to sift out the coin in the least number of turns. (3) Using a balance scale again, what is the smallest number of weights which can be used to find the weight of objects of all integer weights between 0 and 121 grams? (i.e. weigh and object of 1 gram, 2,3,4 and so on) (4) A coin collector received 10 bags of rare coins. One of the bags, however, contains only fake coins. The genuine coins weigh 10 grams each, the counterfeit coins weigh 10.1 grams. Each bag contains 100 coins. The collector has a weighing scale which can weigh any number of coins and give an exact measurement. However, he can use it only once. He is allowed to take any number of coins from each of the bags. How can he determine which bag contains the counterfeit coins?
JaKiri Posted February 18, 2004 Posted February 18, 2004 1, 2 are pretty funky, and I won't spoil them by revealing the answers so quickly. 3, be it not altogether radical, you see obvious repetitive tendencies. 4 I preferred as the poison question.
Radical Edward Posted February 18, 2004 Author Posted February 18, 2004 how is described as a poison question? my favourite is 2. it took me a while to get the answer to that one.
JaKiri Posted February 18, 2004 Posted February 18, 2004 Radical Edward said in post # :how is described as a poison question? Dinner party. Someone reveals that, in one of the glasses of wine not yet served, there's been placed the contents of a vial of poison. Etc.
mooeypoo Posted February 29, 2004 Posted February 29, 2004 I suck at those things but since i had an idea what the heck, I'll give it a try. Worse case scenario - I'll make you laugh Okay, about (2) What I was thinking is that you take the group and devide it to half, putting half on one side of the scale ad half on the other. If you'd know that the coin was HEAVIER or LIGHER you could pick a group, but since you don't know, you need to start (I .. hope? taking halves of each group. [*] You have 2 sides of the scale - one has 6 coins and the other 6 coins, and the scale is not balanced. [*] Now, you take 3 coins of each side and removes them. If the scale REMAINS unbalanced, then your forgery is ON the scale, and you just got rid of 6 coins. If they balance, your forgery is OUT of the scale and you also removed 6 coins. Either way, now you have 6 coins you are SURE are original, and 6 that you can't be certain of. [*] If you put one next to the other on the scale, you'll know if the forgery coin is HEAVIER or LIGHTER than the rest of the coins. Since we already saw the coins on the first "look" (and saw which side is heavier and which is lighter) we now can know which of the 3 coins the fake one can be in. [*] You also know that the other 9 coins are originals. So if you take one DEFINATELY original coin and put it on one side with one coin out of the three you have left, and on the other side of the balance you put the other 2 coins, you can either see right away where your forgery is, or all you need to do is switch coins between the sides, and make sure you switch one coin with the ORIGINAL coin - if nothing happened, it's the coin you didn't touch (the one left on the other side of the balance scale) and if the scale moved, it's the coin you just switched with the original coin. That - of course - depends on the fact you didn't already discover that your forgery coin is the one with the *definate original* one. [/list=1] Err, I hope that made sense, and since I have no idea what you meant by "least number of turns" I just... tried. Hope that if it's not the right answer, it might get close. In any case, that's my try. ~moo
Pinch Paxton Posted February 29, 2004 Posted February 29, 2004 I was given a similar problem at work but it had to do with gold bullion. Pincho.
mooeypoo Posted March 4, 2004 Posted March 4, 2004 Are you going to solve any of those?! (at least tell me if i was right or wrong ...) eek I'm curious. Luckily, curiousity killed the CAT not the COW. Phew ~moo
BrainMan Posted March 16, 2004 Posted March 16, 2004 (1) The fastest way I know is 4. (2) Again, fastest way I can see is 4. (3) The answer is 5. (4) Split the bags into two groups of five. Take five coins from a bag on each side, four coins from another, then 3, 2, 1 [and mark bags based on the pile A or B and the number of coins 1, 2, 3, 4, or 5]. Whichever side is heaver tells you which pile of five bags to look at, and the number of tenths of a gram it is over tells you which bag.
BrainMan Posted March 16, 2004 Posted March 16, 2004 It seems I misunderstood (4) a little. Instead of breaking it up into two piles, you just take 10 coins form one bag, 9 from another, then 8, 7, 6, ect... the number of tenths of a gram it is over will reveal the counterfit.
Pinch Paxton Posted March 16, 2004 Posted March 16, 2004 Did no one get that? I was the only person to get it right at work, but this is a brain - site.
Neurocomp2003 Posted March 16, 2004 Posted March 16, 2004 1) 3 moves 2) 3-4 moves (all the branches were 3 except one 4 branch) 3) 8 coins 4).. 5 and 5 and just keep lifting bags.
BrainMan Posted March 16, 2004 Posted March 16, 2004 For (3), the answer is 5. [i edited my last post to include this answer.] The exact weights you need are {1, 3, 9, 27, 81}.
Neurocomp2003 Posted March 16, 2004 Posted March 16, 2004 BRAINMAN um....how do you get the values 2,5,6,7,8,11.....it says not the values but how many weights for 2 if you need 2 1's then your list will be 1,1,3,9,27,81...to make 8 you would need 1,1,3,3 ,9,27,81 to make 18 you need 9,9 so you list just keeps growing...i think you miss understood the question.
BrainMan Posted March 16, 2004 Posted March 16, 2004 You have a balance, so you can subtract numbers as well as add (by plaing counterweights on the other side of the balance). So fo example, to get 2 you place 3 on one side and 1 on the other. For 5, you place 9 on one side and 3 and 1 on the other. This works for all possible whole numbers. In fact, the general solution for any set of whole numbers you can generate is: for how many numbers you need, (3^N - 1)/2. and the exact numbers needed to generate all numbers is {1. 3, 9, ...3^(n-1)}. In this case, we are given that it must generate numbers up to 121, so we have: (3^N - 1)/2 = 121. Solving for N we have 5. 5 weights are needed to generate all numbers between 1 and 121. Understand now? Sorry if my explination sucks...Im not sure how to explain it better.
Neurocomp2003 Posted March 17, 2004 Posted March 17, 2004 my bad i should read the question better...i was thinking of coins.
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