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anonymous
 5 years ago
You have 26 weights, one of which is lighter than the other 25. How would you determine the light weight in three separate weighings on a balance scale?
anonymous
 5 years ago
You have 26 weights, one of which is lighter than the other 25. How would you determine the light weight in three separate weighings on a balance scale?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if anyone can help me out with this problem by tonight that would be great, I would really appreciate it!

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0I dont know an answer right off hand, but what are your thoughts about it so far?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.03 seperate weighings means that we would have to have 3 seperate groups to weigh in my thinking.. and you only have a 1 in 26 chance of picking the right one.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.026/2 = 13 so pick the side that is lighter and you have 13 to choose from...

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0split that up to 6 and 7; the heavier side is off by the lighter weight choose the 7 to work with

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0or; from the 13 choose one weight not to weigh, if they come out equal, you have the lighter one. If not, choose the lighter side again?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0I can do it in four weighings :)

radar
 5 years ago
Best ResponseYou've already chosen the best response.0What about dividing the weights into three piles, two piles with 9 and one pile with 8 Place the two piles of 9 on the balance scale. Their is only two conditions that can take place 1. The scales are balanced which means the lighter one is in the pile of 8. or 2. The scales are unbalanced. Lets go with situation 2 and remove the lighter (higher side of scale) Now divided these 9 into three piles of three. This would be the second weighing place two piles on three on each side of the scale. If balanced then the lighter weight is in the remaing pile (as before the two conditions are a balance or an unbalance. If they are unbalanced then remove the lighter side and using the third measurement place one on each scale and locate the lighter one. If it were balanced on the second weighing then get the remaining three and place one on each side of the scale and locate the lighter with the third weighing. Now for the earlier case no 1 after the first weighing. Take the eight from that third pile and divide them into a group of three, another group of three, and a group of two. For the second weighing place the two groups of three on the balance scale. This second weighing will either be balanced telling you the lighter weight is in the group of 2, if this is the case, then weigh those two for your third weighing and select the lighter, The case was unbalanced for the second weighing, remove the three from the lighter side, and place two of them on the scales (one each side) and select the lighter one from this final or third weighing.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0got it; i think :) split the load 9  9  8 1) weigh 9  9 and if equal; L is in 8. split 8 to 3  3  2 weigh 3  3, if equal then L is in 2 and weigh 1  1 if 3  3 is not balanced, pick lighter side. weigh 1  1; if equal, L is on the ground, if not equal L is lighter one. 2) weigh 9  9 and not balanced; L is in lighter side split to 3  3  3 weigh 3  3 and if balanced L is in 3 on ground; if not balanced L is in lighter side with (3) either way you have 3 to weigh. weigh 11 if equal L is on ground; if not equal, L is lighter side. All done in 3 weighings :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0It took me awhile, but yeah....i agree with radar :)

radar
 5 years ago
Best ResponseYou've already chosen the best response.0amistre64 as usual you did a very coherent and numerical approach. I was trying to use logic, we both arrived at the same conclusion.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thank you both for your great help!
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