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Mathematics Puzzle: Tipping the Scales?
A balance scale sits on a teacher's table, currently tipped to the right. There is a set of weights on the scales and on each weight is the name of at least one pupil.
On entering the classroom, each pupil moves all the weights carrying his or her name to the opposite side of the scale.
Prove that there is *some* set of pupils that you, the teacher, can let in which will tip the scales to the left.
To clarify:
Scales begin tipped to the right but not necessarily with all weights on the right.
All weights are not necessarily equal.
Pupil 1 enters, moves the weights with his or her name to the *other* side, pupil 2 enters and does the same with his or her weight, which may include some that pupil 1 moved etc.
2 Answers
- Barry GLv 77 years agoFavorite Answer
Very good puzzle.
Zak has not understood the difficulties in the conditions of the question. (1) We are not told how many weights or what values there are, nor which side of the scale they are at the start. (2) There can be more than one pupil's name on each weight. (3) Each pupil moves ALL of the weights bearing his/her name to the OPPOSITE side of the scale.
The last condition means that letting in all of the pupils could result in the scale ending up tipped to the right. For example, if there are 2 pupils and all of the weights bear both of their names, then the 1st pupil will transfer all of the weights to the opposite sides, but the 2nd pupil will transfer them all back to the original sides, leaving the scale loaded as at the start and therefore tipped down to the right.
We are not told the weights being used, nor what combinations will tip the scales to the left. I think this means that, in order to be sure that the scale tips to the left we could ensure that either (a) ALL of the weights end up on the left side of the scale, or (b) all of the weights end up on the opposite side of the scale from which they started. However, while either condition is SUFFICIENT to ensure that the scale tips to the left, neither is NECESSARY to ensure this happens, because if one weight is heavier than all of the others put together, then the position of that one weight alone is sufficient to determine the outcome, regardless of what happens to the other weights. This distinction between necessary and sufficient conditions means that it is more difficult to disprove the theorem by finding a counter-example.
For example, suppose there are 3 weights (A,B,C) which all start on the right, and two pupils (1,2). If the weights are labelled (1) (2) and (1,2) then it is obvious that regardless of whether one or both pupils enter the classroom, one weight.will remain on the right. So we cannot ensure that all of the weights are transferred to the opposite side (Condition b). However, if either A+C>B or B+C>A or A+B>C then the scale can be made to tip to the left. Regardless of the values of the 3 weights, one of these 3 inequalities must be true - at least one (eg the lightest of the 3) must be lighter than the sum of the other two - so the scale can be made to tip to the left.
- 7 years ago
If there are a certain amount of weights tipping the balance scale to the right, by moving them left - as the pupils walk in - then the scale will tip left.
