8 bags of gold problem

That group of four bags has the lighter bag. When that number of additional coins is multiplied by 10, the bag on the heavier side which 8 bags of gold problem that many coins on the scale is the counterfeit bag. So each possibility has a unique scenario.

It can be done in 2 weighings The first weighing compares 3 and 3. Reserve one group and put the other two groups onto the balance scale. Since the king owned all of the gold in his country, it was obvious that one of the eight people he trusted was cheating him.

There are nine balls and a balance not a scale One of the balls is lighter than the rest How do you find the lighter ball by using the balance only twice How do you do it?

If they are the same, the counterfeit is one of the reserved group. But again, perhaps another answer may be enlightening on that. General solution 12 coins problem Below, you will find my general solution to the 12 coins problem.

Otherwise, the lighter pan has the counterfeit. If they do not balance, then 11 is the good pill. However, one cannot answer your question without further details since we clearly do not know how much of each object we have and how heavy it is.

The resulting distribution of coins over scales is the one given at the beginning. So all we have to do is to distribute the 12 coins over the scales of the three measurements in such a way that no coin participates in the three measurements in the same way or mirrored as any other coin.

Now to the side which weighs less so we know all of those coins are authenticadd as many coins from any bag on that side as is necessary to balance the scale.

SOLUTION FOR GOLD COINS

How much does one plastic bag weigh? Since Step 1 will tell you which group of 3 the light egg is in So the scale will read 3 grams. A farmer has 9 eggs and thinks one is ligher than the other 8 if he used a balance scale to weigh the eggs only 2 different times how could he find the one egg that weighs less?

Now, in order to proceed, we must keep track of which side is heavy for each of the following weighings. Thanks for the help Date: Now weigh 9,10,11 against 1,2,3.

The amount of the difference in weight determines which bag is counterfeit.

SOLUTION FOR THE GOLD COINS PROBLEM

The table can now be reduced to mapping 12 statements about coins being heavier to 12 outcomes each outcome taken from a different pair. I know there is no scheme with fewer weighings that will work because there is no other way to narrow eight down in smaller groupings.

The scale will show which side is heavier and it is that side which has the counterfeit coin s. If one… group is lighter that side of the balance is higherit's in that group. We know that the gold coins must be in one of those bags, so there are three possibilities when we weigh the three coins we removed: If they are the same, the counterfeit is one of the reserved group.

So the scale will read 3. If the scale 8 bags of gold problem - you know the light egg is on the "light" side. It is a systematic and rather elegant approach in my humble view. On special occasions he asked them to bring the bags back so he could look at them.

Let's simplify the problem and say there are only 3 bags each with 2 coins in them. Divide the balls into three groups of three. If our electric scale is replaced by a scale of libra, I don't believe it would be possible to answer this puzzle with only one measurement of weight.

Since the numbering of the coins is irrelevant, the only remaining choice we have is which outcome of each pair we will use in the table. His court mathematician thought that it could be done in fewer weighings. If you think that the king cannot find the lighter bag in fewer than three weighings, prove it.

What do you think? For instance, if the three results are: Place the bags 2, 4, 6, 8, and 10 on the other side of the scale.

So the weight of the 3 coins on the scale are all 1 gram. Eleven are of equal weight.Eight Bags of Gold Problem Statement: There was a king who gathered up all the gold in his land and put it into eight bags. He made sure that each bag weighed exactly the same amount.

The king then chose eight people in his country whom he trusted the most, and gave a bag of gold to each of them to keep safe for him. Luckily, your goldno.8 bag changes, too. Design your first bag, then update your straps and accessories as often as you wish. We sell all our parts separately, so you don’t have to invest in an entirely new bag to achieve a whole new look.

Eight Bags of Gold: Matthew for Children (Arch Books) by Janice Kramer. Format: Paperback Change. Write a review. See All Buying Options. There was a problem filtering reviews right now. Please try again later. out of 5 stars EIGHT BAGS OF GOLD.

Problem statement. A wealthy king has 8 bags of gold that gives to some of his most trusted friends. All the bags have the same weight and the same amount of coins in the bags is all of the gold in the kingdom. Although, the king herd that a local woman received a.

Apr 17,  · Now, if 1,5,6 vs 2,7,8 does not balance, and 2,7,8 is the heavy side, then either 7 or 8 is a good, heavy pill, or 1 is a good, light pill. For the third weighing, weigh 7 against 8. Whichever side is heavy is the good pill.

Jan 12,  · In this case, take 2 pieces from Bag #1, and 1 piece from Bag #2 and put them all on the scale together. If the scale shows a weight of 3 lbs, then Bag #3 contains the real gold. If the scale shows a weight of lbs, then Bag #2 contains the real dominicgaudious.net: Resolved.

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8 bags of gold problem
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