ANSWER TO PUZZLE-3

 

The approach to solving this problem is given in the HINT; as mentioned there, let us choose the first weight as 1 to start with. We need to choose the next weight as far away from 1 as possible , but simultaneously ensuring that this choice ensures the possibility of weighing all the integers up to that number. Such a number is 3. By choosing 3 along with 1, we find that we can weigh 1,2,3 and 4 in single weighing.

The critical question is – What should be the next weight as that would by default decide the fourth weight also. Let us incrementally look at weights after 3 to see how far away a number we can choose and still ensure that all the interim integers are covered. We can take 4 as the next weight which will take us up to 1,2,3,4,5,6,7and,8.

But should we not try the next number which perhaps may take us a farther distance. Yes, we can look at 5 instead of 4-{1,3 and 5}. This will allow us to weigh 1,2,3,4,5,6,7,8,9.

Now let us discard 5 and look at 6 which will allow us {1,3 and 6}to weigh 1,2,3,4,5,6,7,8,9 and 10.

We can keep trying like this and when we reach 9 ,{1,3 and 9}we find that we can weigh up to 13.The balance weight will be 27 and we find that with that weight we are able to cover all the way up to 40.

What happens if we skip 9, and choose 10{1,3 and 10}. We find that we can weigh 1,2,3,4 , 6,7,8,9,10,11,12,13 and 14 but we cannot weigh 5 in one weighing.

Thus the answer is 1,3,9 and 27.

Further questions: What is the next number we need to choose so that we can get all integers without break? We find it is 81 and with that we can weigh up to 121 in one weighing. Isn’t this interesting?

Thus the series 1,3,9,27,81,…. will be provide to all integers up to the sum of all the numbers in the series, through a combination of addition and subtraction of all or some of the numbers.

I will request the readers who find this puzzle interesting to attempt PUZZLE-8 and try to figure out the connection between these two puzzles.


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