rookie1ja 11 Report post Posted March 30, 2007 Weighing VIII. - Back to the Water and Weighing Puzzles Suppose that the objects to be weighed may range from 1 to 121 pounds at 1-pound intervals: 1, 2, 3,..., 119, 120 and 121. After placing one such weight on either of two weighing pans of a pair of scales, one or more precalibrated weights are then placed in either or both pans until a balance is achieved, thus determining the weight of the object. If the relative positions of the lever, fulcrum, and pans may not be changed, and if one may not add to the initial set of precalibrated weights, what is the minimum number of such weights that would be sufficient to bring into balance any of the 121 possible objects? This old topic is locked since it was answered many times. You can check solution in the Spoiler below. Pls visit New Puzzles section to see always fresh brain teasers. Weighing VIII. - solution There are necessary at least 5 weights to bring into balance any of the 121 possible objects. And they weigh as follows: 1, 3, 9, 27, 81g. Suppose that the objects to be weighed may range from 1 to 121 pounds at 1-pound intervals: 1, 2, 3,..., 119, 120, 121. After placing one such weight on either of two weighing pans of a pair of scales, one or more precalibrated weights are then placed in either or both pans until a balance is achieved, thus determining the weight of the object. If the relative positions of the lever, fulcrum, and pans may not be changed, and if one may not add to the initial set of precalibrated weights, what is the minimum number of such weights that would be sufficient to bring into balance any of the 121 possible objects? Share this post Link to post Share on other sites

Guest Report post Posted July 3, 2007 Why isn't the answer "3". If you put the 5 pound weight on the scale, you only need the 4 and 1 to balance it out. Perhaps I am missing what you are asking. Share this post Link to post Share on other sites

Guest Report post Posted October 19, 2007 Your solution would work if it weren't for the initial weight that was placed on the scales. Think of it this way: you know there are 121 weights, but you're not allowed to see them. The person in charge of the weights randomly selects a weight and places it on the empty scale, but you have to be able to balance the scales from a preselected (or "precalibrated") set of weights (within the original 121) that you've removed from the original lot. With the set of 1, 3, 9, 27, and 81, you can balance the scales no matter which weight the person picked. For example, if he places the 2 pound weight on pan A, you can add the 1 pound weight to pan A and the 3 pound weight to pan B, and the scales will be balanced (1+2 = 3). Share this post Link to post Share on other sites

Guest Report post Posted October 22, 2007 You would need at least 5 weights: 1, 3, 9, 27, and 81 pounds. With one of each, using addition and subtraction, all values from 1 to 121 can be made. I remember hearing one like this long ago. It was about a 40 pound rock that split in 4 pieces of different sizes in such a way, that all values from 1 to 40 pounds (in 1 pound intervals) could be formed on a set of scales. Find the weights of the 4 pieces. Of course, the solution there was: 1, 3, 9, 27. Alternatives could be 6 pieces weighing 364 pounds (last piece 243); 7 pieces weighing 1.093 pounds (last piece 729); or 8 pieces weighing 3.280 pounds (last piece 2187). And so on, and so forth... Happy Puzzling, BreakingRock Share this post Link to post Share on other sites

Guest Report post Posted February 6, 2008 I think you only need 1 1 pound weight. 1) Placing 1 pound measure on one side, find the 1-pound item 2) Mark the item '1 pound' 3) Place both 1-pound items on one side and find the 2-pound item 4) Mark it '2 pounds' 5) Remove the 1 pound item and add the 2 pound item to find thr 3 pound item. 6) Continue this process -> Remove the second highest found item and replace it with the highest -> Find the new highest, label and repeat. You could do this with all 121 items and successfully identify all of them. Share this post Link to post Share on other sites

Guest Report post Posted February 25, 2008 if you keep all the weights on one side of the scale, you need 9 different weights. 1, 2, 2, 6, 11, 11, 33, 55 Share this post Link to post Share on other sites

rookie1ja 11 Report post Posted February 25, 2008 if you keep all the weights on one side of the scale, you need 9 different weights. 1, 2, 2, 6, 11, 11, 33, 55 the question was: What is the minimum number of such weights that would be sufficient to bring into balance any of the 121 possible objects? I used 5 weights, you used 8 (separate objects). Share this post Link to post Share on other sites

Guest Report post Posted March 17, 2008 This solution is the basis (or is based on, take your pick) of a number system called "balanced ternary". Instead of the "digits" 0, 1 and 2 that you have for the ordinary ternary number system you have the "digits" -1, 0, +1 (frequently represented as "-", "0" and "+"). It has the interesting property that you don't need any special conventions (such as a "-" at the beginning of the number) to represent negative numbers. To negate a number you replace the +1s with -1s and vice versa, so since "5" is "+--", "-5" will be "-++". The signs "just work" in adding, subtracting and multiplying. Share this post Link to post Share on other sites

Guest Report post Posted March 20, 2008 rookie1ja - I know I've asked you this on one of your puzzles before but could you explain the math behind finding the answer to this puzzle. I assume there is some formula behind it. Share this post Link to post Share on other sites

rookie1ja 11 Report post Posted March 20, 2008 rookie1ja - I know I've asked you this on one of your puzzles before but could you explain the math behind finding the answer to this puzzle. I assume there is some formula behind it. I would like to see the formula as well ... I just added the first to the next one and it all went quite logically ... I knew that I had to start with 1 ... then two would be measured with 1 to the same pane and 3 on the other pane ... so I had 1 and 3 ... with those two I could measure three and four so I could not measure five, but since I already could measure four I could use it on the same pane and have 9 on the other pane ... thus having 1, 3 and 9 ... and so on btw, notice the multiplication by 3 Share this post Link to post Share on other sites

Guest Report post Posted March 22, 2008 Here's another way of explaining how to find the solution: Start with 1. Sum the total of your weights + the next number you need. Repeat. Example: 1 Total of your weights: 1 Next number you need: 2 1+2=3 1,3 Total of your weights: 4 Next number you need: 5 4+5=9 1,3,9 ... if you keep all the weights on one side of the scale, you need 9 different weights. 1, 2, 2, 6, 11, 11, 33, 55 You said 9, you listed 8, and you only need 7: 1,2,4,8,16,32,64 Share this post Link to post Share on other sites

Guest Report post Posted April 24, 2008 each number is sum of all previous numbers in the sequence * 2 + 1 until sum is > = 121 1, 2*1 +1 = 3, 2 * (1+3) + 1 = 9, 2*(1+3+9)+1 = 27, 2*(1+3+9+27) + 1 = 81 Share this post Link to post Share on other sites

Guest Report post Posted July 14, 2008 We're talking about Ordinals here. What numbers (weights) do you need to add up to every number between 1 and 121? The answer is 7: 1, 2, 4, 8, 12, 32, & 64 Share this post Link to post Share on other sites