There are 8 identical balls, one weighs less: can you find the lightest one? The solution to the riddle

In the’8 ball riddle the situation is this: 7 balls they have it same identical weight, While one weighs 1 kilo less of the others. We have a scale with two plates available, but we can only perform two weighings. What is the correct way to arrange the balls in our two weighings to ensure we find the ball that weighs less?

Unlike the riddle of the 12 coins, in this question there are no more cases to consider, but rather the solution is unique.

If we decided to divide the 8 balls into two groups of 4 and weigh them, we would immediately identify a group of 4 balls which – weighing less – will contain the ball we are looking for. At this point, however, it will be impossible to find our solution with just one more weighing: if we divide them again into two groups of 2 and choose the one that weighs less, we will have identified a pair where the ball with the least weight will certainly be, but we won’t be sure of identifying which one it is! To do this, a third weighing would be necessary, which however we do not have available.

What then is the solution to the riddle of the 8 balls? Here it is. We have to choose 6 balls randomly and divide them into two groups of 3leaving the 2 remaining balls aside. So let’s do our first weighing by placing 3 balls on one side and 3 on the other, and here there are two possible cases:

1. If the two groups of 3 weigh exactly the samethen surely the lightest ball is not among the 6 chosen balls, but rather will be in the 2 balls that we have put aside. At this point we do our second weighing by comparing these two balls, one on each plate, and the ball that weighs less will be exactly the ball we were looking for.
2. If the two groups of 3 do not have the same weightthen our ball will be among the 3 balls that weigh the least in our first weighing. At this point, let’s randomly take 2 balls from the group that weighs less and compare them in the second weighing: if they weigh the same, the ball that weighs less is precisely the one we didn’t weigh; if, however, one of the two weighs less, it will be the ball we were looking for.