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The minimum number of weighings required is 1. First I'll explain how to get the answer then the logic required to get that answer.
1. What to do. Take 1 pearl from the first bag,3 pearls from the second bag, 9 pearls fromt the third, 27 pearls from the fourth and 81 pearls from the fifth. Weigh this and you'll get a number. Using only this number you can figure out which bags contain which types of pearls.
2. The logic. Fistly look back at the problem trickier pearls and look at the number system used. As you may notice the number of perals taken from each bag doubles as you move to higher bags. Another thing to notice is the two different weights of the pearls. (9 and 10 grams).In this question I took 1, 3, 9 ,27, 81 pearls. You may notice that these are all multiples of 3 but you may also notice in this question we were dealing with three different weights(9,10 and 11grams). So one thing to realize is that if every bag could weigh 9,10,11 and 12 grams you would have to take 1, 4, 16, 64, 256 pearls bag by bag. The reason you have to do this? The reason you have to take these varying amounts is so that no matter which bags contain which types of pearls no other combination of pearls will add up to that weight. The cause of this is the way the number system works. Now in our question the weight of the pearls from each bag will have a minimum and maximum variance. Lets set up a chart to show this.
------------------Bag---1---2---3----4----5
weight if fake pearls 9--27--81--243--729
weight if blue pearls 10--30--90--270--810
weight if black pearls 11--33--99--297--891
So now I'm going to find some of the maximum variances. For the first bag the maximum variance is 2(This is because the weight of the bag can vary within 2) For bag 2 the maximum variance is 6. For bag 3,18. For bag 4,54. Now for bag 5 I will show you the minimum variance, which is 81. This is because the weight of the bag can vary by at least 81. Now the reason we showed maximum variances was to show how much the weight of the pearls you weighed could vary due to that one bag. Now to answer the question "Why can't any two pearl bag combinations add up to the same weight when weighed in this way" Well if we add up the maximum variances of bags 1 through 4(2+6+18+54) we get 80. Now as we can all see this number is 1 below the minimum variance of the bag that comes next(bag 5) So this shows that no matter what each bag contains it will never vary enough to cancel out a change in a higher bag. This is why every bag/pearl combination weight will be unique. For all those who did and didn't understand this you may want to read this just to make sure you understand properly.
1. This will work for any amount of bags.
2. How you draw pearls from the bags(multiples of2,3,4,etc.)is not dependent on the different weights involved but it is dependent on the maximum variances. |
