You give simple problems to avid puzzlers and mathematicians, and you always find some twists to the problem that you never thought of. This puzzle was one of them. But before I go into the complications, here are the people who responded with correct answers:
- Amit Mittal, Kaushik Paul, Deepak Yadav and Prakhar Prakash with partially correct answers i.e. they found one correct answer each
- Suman Saraf (and his daughter), Pratik Poddar, Karan Sharma, Sid Mulherkar and Abhishek Masand gave both the correct answers
There are two answers to the puzzle
When I did the question, I did it more intuitively (and also trial and error and so did some others). I am reproducing Karan’s logic to the problem here:
Two observations that we can make after looking at the weights:
1. All weights are even or odd
2. No weight is repeated
Using brute force further, i observed two possible cases-
1,5,7,9 and 2,4,6,10
Sid Mulherkar, however, very rightly pointed out that we made an assumption that the weights are integers. The problem does not say that, and theoretically therefore, the numbers could even be 1.5, or 5.5 etc. It just happens to be that the two solutions are both integers only. Reproducing his full solution (very similar to Pratik Poddar as well)
Let the measures of the weights be a,b,c,d and WLOG let a<b<c<d (where < means less than or equal to).
Now using a<b<c<d, we can only have two possibilities. They are:
In case a) we get the following system
Subtracting (ii) from (iii) and adding the resulting equation to (i) we get b=4. From here we can easily calculate value a,c,d. giving us (2,4,6,8)
In case b) we get a similar system as case a) but this time we have
Subtracting (ii) from (iv) and adding this to (i) we get b=5. Then we easily get a,c,d. The final answer is (1,5,7,9).
Hope you all enjoyed the puzzle!