Once there was a king who took all the gold in his land and put it into 8 bags. he made sure that each bag?
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weighed exactly the same amount. The king then chose the 8 people in his country whom he trusted most, and gave a bag of gold to each of them to keep safe. On special occasions he asked them to bring the bags back so he could look at them. one day the king heard from a foreign trader that someone from the king's country had given the trader some gold in exchange for some merchandise.The trader coundn't describe the person who gave her the gold, but she knew they were from the king's country. One of the 8 people he trusted was cheating him. the only scale in the country was a pan balance. the scale won't tell what something weighed, but could compare 2 things and determine which was lighter and which was heavier. the king asked the 8 to bring their bags. the king only wanted to use the balance 3 times. his court mathematician said it could be done in less. what do you think? 1.develop a way for comparing bags tht will always find the light one 2.explain how you're sure it's right: prove
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Answer:
weigh 3 in one and 3 in the other: if they are equal: weigh each of the other 2 in one pan each. One will be lighter, voila If they are different: take the lighter set of 3, weigh one in one pan and one in the other and hold one aside. If one is lighter, voila, if they are equal it's the one set aside. that's the whole thing in 2 weighings
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Other answers
Forget the balance and put them all in a bathtub full of water one at a time. The one that causes the water to rise a lower level then the others is the bag with the missing gold. And that took 0 uses of the balance so i win.
outbaksean
separate them into 2 groups of four, then take the lighter one and split that in half keep doing the same thing until u get the light one.
a,b,c,d,e,f,g,h Take one half and put it on the right, then the other half and put it on the left (a,b,c,d) : (e,f,g,h) Now take the lighter side and do the same thing, half on the left and half on the right. (a,b) : (c,d) Take the lighter side and through the other one who was carrying it, into the dungeon. a : b Only one problem, i hope the king doesn't get the bags mixed up or he just through the wrong person into the dungeon.
Sherman81
Ronin's right. The other answer requires 3 weighings (8 = 2^3)
Philo
Ronin is correct. But it is worth noticing that his method works for 9 bags, as well as 8. If ABC weigh the same as DEF, weigh G against H, and if they weigh the same, then it must be the 9th bag, I, that we seek. Generalising, one light bag in 27 bags can be found in just 3 weighings, one light bag in 81 in just 4 weighings, one light bag in 243 in just 5 weighings. And just n weighings are needed for 3^n bags. The method is identical, weigh one-third against one-third and leave one-third on one side. This identifies which one-third needs to be investigated further, then divide that into one-third weighed against one-third, with one third left aside. Then proceed until you reach the last three bags, and weigh one against a second one, and leave one on one side. Of course if 2 or more bag-holders have cheated the King, this method will only catch one of them, at best. What could happen is that in the last weighing, one light bag is in one scale, the other light bag is in the other scale, and that they are light by exactly the same amount and the scale doesn't tip. And then an innocent bag-holder is condemned on the assumption that the third bag must be lighter, when in fact it is heavier!
brucebirchall
in three steps here it is 1 ) seprate in 2 groups(in one group there r 4 bags) and weight it. 2 )ONE side will be lighter now the bags which are on lighter side make divide them in 2 groups(in one group there r 2 bags) and weight them one side will be lighter. 3 )Now taking bags of lighter side put one bag on one pan and second bag on seond pan and weight it now king can figure out who is cheating
Wolverine
Ronin is correct.
River Rat
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