How can you find the heavist of eight marbles in only two weighings?

Answer 1

well this will be hard if the other 7 marbles have different weights. my answer will only rely on the condition that out of 8 marbles, 7 marbles have identical weights and only 1 marble out of the 8 is significantly heavier. Also, that the marbles are weighed here on earth. This is what to do (1) divide the marbles into 3 groups with 2 groups having 3 marbles each and 1 group having only 2. example: OOO + OOO + OO (2) weigh against each other the groups having 3 marbles each (1st weighing) OOO vs OOO (3) at this point there will be two possible outcomes - one is that the marbles will be balanced (1st outcome); and two is that one side of the balance will fall lower than the other (2nd outcome). for instance: 1st outcome - OOO = OOO 2nd outcome - OOO > OOO or OOO < OOO (4) if the result of the 1st weighing is the 1st outcome then just weigh the remaining 2 marbles for the second weighing then it will be easy to know which one is heavier. (5) if the result of the 1st weighing is the 2nd outcome then choose the heavier group which is composed of 3 marbles (forget about the lighter group and the group composed of 2 marbles). Now from this group, choose 2 out of the 3 marbles and weigh them against each other and this will be the 2nd weighing. i.e. O + O and O to be set aside Again 2 possible outcomes here for the 2nd weighing - 1st outcome is that the 2 marbles which were selected do not balance each other, therefore, the one that falls lower in the balance is the heaviest marble; i.e. O > O and the 2nd outcome, is that the 2 marbles balanced each other which will lead to the conclusion that the marble that was left out during the selection process is the heaviest marble. i.e. O = O thus the remaining O is the heaviest.