I have 13 identical balls that weigh the same, and out of them, one is an odd one weighing less or more. How can I find the odd one out with a simple balance? What is the least number of times I require a simple balance to find the odd one out?
We can start with basic iteration to weigh all the balls against each other, but that would take too many turns. Instead, we can group them in fours to reduce the iterations and maximize the result derivation vector. In the first iteration, we can narrow down to minimum of eight, in the second to minimum of three, and in the third to one. Therefore, the minimum number of turns required is three.
Below is an instruction that describes a task, paired with an input that provides further context. Write a response that appropriately completes the request.
### Instruction: I have 13 identical balls that weigh the same, and out of them, one is an odd one weighing less or more. How can I find the odd one out with a simple balance? What is the least number of times I require a simple balance to find the odd one out?
### Input: 
### Response: We can start with basic iteration to weigh all the balls against each other, but that would take too many turns. Instead, we can group them in fours to reduce the iterations and maximize the result derivation vector. In the first iteration, we can narrow down to minimum of eight, in the second to minimum of three, and in the third to one. Therefore, the minimum number of turns required is three.