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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?

RC Wong
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  • You are thinking of weighing with a two-pan balance, but the question is referring to a scale that displays the weight. – saulspatz Jul 05 '18 at 15:44
  • Weigh one from pile $1$, two from $2$, ..., $2016$ from the last. – lulu Jul 05 '18 at 15:44
  • The problem asks for the least number of times one might have to weigh a ball. If you get lucky, then the first weighed ball is 2017 mg and you're done. – Narlin Jul 05 '18 at 15:47
  • its possible to get the result always with just one use – Asinomás Jul 05 '18 at 15:48
  • Hi, thanks for clarifying this. I got too used to 2-pan scale problems that I missed that it was asking the use of only 1 scale. So logically, 1 weighing AT THE LEAST would be needed to find the balls with 2017 mg weight. So for this type of problem there wouldn't be any solution? Thanks again – RC Wong Jul 06 '18 at 08:13

1 Answers1

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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 $k$.

Asinomás
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  • I got the weight of such would be $\frac{2016^2.2017 - 2015k}{2}$. If all of them were to be 2016mg then you would get $\frac{2016^2.2017}{2}$ and thus the difference between them is $\frac{2015k}{2}$ and you can find k from that. Would you want to show me how you got $2016\times 2016+k$ – Satish Ramanathan Jul 05 '18 at 16:19
  • @SatishRamanathan you are correct – Asinomás Jul 05 '18 at 16:25