Okay, I've done those 'weighing puzzles', but this one stumps me as there seems to be no solution to it... Google gave me wrong solutions, as there were eventualities that seemed like nobody considered... Anyway, here it is ;)
There are 12 oranges, all but one have the same mass (or weight). When it comes to the last orange, we don't know if it is heavier or lighter than the rest. How can we get the 'odd' orange out, using only thrice, a balance.
And by balance, I mean this sort of balance without any scale:
http://www.lenovoblogs.com/designmat...04/balance.jpg
I tried dividing into 3 groups of 4.
1 2 3 4 | 5 6 7 8 | 9 10 11 12
First try: taking first two groups:
- If they balance, the odd orange is in the last group of 4.
- If not, it's among the eight used.
Taking the worst scenario, they are in the 8.
Divide into 4 more groups: 1 2 | 3 4 | 5 6 | 7 8
Second try: take 1st and 2nd:
- If they balance, it's between the 4 left.
- If not, they are between the first 4.
And with one remaining try, I don't see how I can get the odd orange from 4 oranges :(
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I tried taking 4 groups of 3.
1 2 3 | 4 5 6 | 7 8 9 | 10 11 12
First try: taking first two groups:
- If they balance, the odd orange is in the first 6.
- If not, it's among the last 6.
Well, I suppose now dividing into groups of 2.
1 2 | 3 4 | 5 6
Second try: taking first two:
- If they balance, the odd orange is in the last group of 2.
- If not, it's among the 4 used.
And worse case scenario, I can't get the orange from 4...
So, the aim is to get down to 2 oranges with one last weighing. So that on the last try, we take one orange at random from the 'good' oranges and compare with any of the two:
- If they balance, the one not picked is the odd orange.
- If they don't, the one picked is the odd orange.
Any idea? :)