Here's the problem that we're trying to solve:
There are 2016 piles of steel balls with the same appearance. In each pile, there are 2016 balls. It is known that among them, 2015 piles of balls are quality products and 1 pile of balls are defective products. A quality ball weighs 2016 mg each and a defective ball weighs 2017 mg each. How many times at least must the balls be weighed by a scale until the pile of the defective balls can be found?
The wording of the problem seems a bit confusing but I assumed for the first weighing you can split the piles into 3 (672 piles each). If the first 2 piles weighed make the scales balanced, then the third pile is the one with the defective balls, otherwise the heavier pile would be the pile that has the defective balle. Then, split the 672 piles into 3's again and repeat process. That would leave 224 piles which you split into 2, taking the heavier pile, and so on. I got around 8 to 9 weighings before finding the defective pile.
However, the answer key states that only 1 weighing is needed. Am I not seeing something?
Add $i$ balls from the $i$'th pile and then weigh them all at the same time.
The weight will be a number of the form $\frac{2016\times2017}{2}\times 2016 + k$ where $k$ is a number between $1$ and $2016$.
Then we know that the defective pile is pile $k$.