9 weights
Problem
First Puzzle:
Can you always work out which weight is heavier in just two weighings of the balance?
Could you always work out which is heavier in just two weighings if you had been given 10 weights?
Second Puzzle:
You have now been given 27 weights, one of which is slightly heavier than the rest.
Can you always work out which weight is heavier in just three weighings of the balance?
Third Puzzle:
You now have 9 weights again, and know that one is a slightly different weight, but you don't know if it is heavier or lighter.
Can you explain how you would always be able to tell which weight is different, and whether it is heavier or ligther, in just three weighings ?
Could you always work out which is the odd one out in just three weighings if you had been given 10 weights?
A very challenging follow-up puzzle that you might like to try is The Great Weights Puzzle.
Getting Started
For the first part of the question, what can you say about the weights on the platform that goes down?
If it balances, what can you say about the weights that haven't been placed on the balance?
For the last part, try to ensure that whatever the outcome of a weighing, you learn a lot about the weights.
Student Solutions
Patrick from Woodbridge School gave a very clear explanation of how to work out which weight is heavier, in just two weighings of the balance, when you have been given 9 weights and you are told that one is slightly heavier than the rest:
It is always possible to find the different weight, using a simple method.First, we think, "If we only had 2 weights, would this be possible in one weighing?". The answer is yes.
Then, "If we had three weights and one weighing, is this possible?" The answer is still yes, since we can put one weight on both sides of the balance. If they are equal then the unused weight is heavier, and if they are unbalanced then the side that goes down holds the heavier weight. Let us call this method Method 1.
We can divide the nine weights into three now:
1, 2, 3
4, 5, 6
7, 8, 9
BALANCE 456 and 789
If the weighings are equal then we can use Method 1 on weights 1, 2 and 3 since we know that all the other weights are equal so they cannot contain the heavier weight.
If the weights 4, 5 and 6 go down on the balance then we can use Method 1 on them since 4, 5 and 6 must contain the heavier weight.
If the weights 7, 8 and 9 go down then we can use Method 1 on them since 7, 8 and 9 must contain the heavier weight.
Thomas from Wolsey House Primary School, Maccy from Limavady Grammar School, Esther, Alex from Hassall Grove P.S. in Australia, Lucy from Flegg High School and Harry from Dumpton School in Dorset all gave very similar explanations of how to solve this problem. Here is Harry's account:
With 9 weights, put 3 on the left scale, 3 on the right scale, and 3 on the side. If scales balance, odd weight must be on the side, otherwise it must be on heavier side of scales. Take the group of 3 weights including the odd one. Weigh 2 of those weights. If scales balance, the one on the side is the odd one. If not, the heavier weight on the scales is odd.
Daniel from Barclay School then built on this strategy to work out which weight is heavier in just three weighings of the balance when you have been given 27 weights and been told that one is slightly heavier than the rest:
Teachers' Resources
This problem puts emphasis on trying to gain as much information as possible from each weighing, whatever the outcome. So if a weighing has one outcome which gives you a lot of information, but another outcome which doesn't give you much new information, it is probably not going to be a useful way of identifying the odd weight.
This problem can be extended further by asking how many weights can be sorted in 3 weighings, 4 weighings and more generally n weighings, when you know one is heavier.
To do this, pupils should first try to spot the pattern, then try to explain why this works by looking at what proportion of weights can be discarded at each weighing.
What happens if two of the weights are heavier than the rest? What is the minimum number of weighings now needed to guarantee being able to identify the two heavier weights
A more challenging follow-up problem can be found at The Great Weights Puzzle