# Solving the Weighing Balls Problem in One Weighing | Precise Scale Method

• Robb
In summary, the conversation discusses a method for determining a defective box of balls that weigh either one ounce too heavy or one ounce too light. By using a precise scale and selecting a specific number of balls from each box, one can determine which box is defective and whether its balls are too heavy or too light in just one weighing. The conversation also addresses the number of balls being weighed and the expected outcome, as well as a specific scenario where all the balls in one box are one ounce too heavy. Finally, the conversation concludes with a clarification on the method and the attempt at a solution. Overall, the conversation provides helpful insights and strategies for approaching the problem.
Robb

## Homework Statement

Consider a set of boxes, each containing 20 balls. Suppose every ball weighs one pound, except that the balls in one box are all one ounce too heavy or one ounce too light. A precise scale is available that can weigh to the nearest ounce (not a balance scale). By selecting some balls to place on the scale, explain how to determine in one weighing which is the defective box and whether its balls are too heavy or too light.

## The Attempt at a Solution

Take one ball from the first box, two from the second box, three from the third etc.
Hence, Σn, from n=1 to 20. I'm not sure how to determine which is the defective box and whether it's too heavy or too light. Please advise.

How many balls are weighed ? What are possible outcomes ?

210 balls are weighed and they either weigh 3,360lbs and 1oz or 3359lbs and 15oz

Sorry, don't know what I was thinking there. The balls weigh greater that 210lbs if the defective balls are too heavy and less than 210lbs if they are too light.

You seem to know there are 20 boxes. Where did you get that information ? It's not in the problem statement.
210 * 20 = 4200 , why do you expect 3360 ##\pm## 1 ounce ?

BvU said:
You seem to know there are 20 boxes. Where did you get that information ? It's not in the problem statement.
210 * 20 = 4200 , why do you expect 3360 ##\pm## 1 ounce ?
Sorry, looks like I left that info out. So, if I sum from n=1 to 20 I get 210 balls which means their weight is greater than 210lbs if the defective balls are too heavy and less than 210lbs if they are too light.

Now suppose ALL the balls in box 8 are 1 ounce too heavy. What would the 'precise scale' say ?

8.5lbs

sorry, 210lbs 8oz

My mistake for writing 210 * 20. Should have been 210 * 1. Never mind.

Now we turn it around: if the precise scale reads 209 pounds 3 ounces, which box is suspect ?

20

?

actually, 1

How do you deduce that ?

ok, I think it's 13 because the scale is 13oz shy of 210lbs.

Right. Do you get the picture now ?

Am I correct in assuming that I don't need to figure out the actual defective box but rather a method for determining that box?

Robb said:
I'm not sure how to determine which is the defective box and whether it's too heavy or too light
And I suspect you can now also describe the method.

PS was the
Robb said:
Take one ball from the first box, two from the second box, three from the third etc.
your attempt at solution or was it in the problem description ?

BvU said:
And I suspect you can now also describe the method.

PS was the
your attempt at solution or was it in the problem description ?

Probably a little of both. I appreciate you taking the time! I'd buy you a draft if I could:)

berkeman and BvU

## 1. How does the precise scale method solve the weighing balls problem in one weighing?

The precise scale method uses a mathematical approach to determine the weight of each individual ball by using a precise scale and a specific formula. This allows for all the balls to be weighed in one weighing, rather than multiple weighings.

## 2. What is the formula used in the precise scale method?

The formula used in the precise scale method is based on the concept of finding the average weight of all the balls and then subtracting the weight of one ball from that average. This gives the weight of the remaining balls, which can then be used to determine the weight of each individual ball.

## 3. Can the precise scale method be applied to any number of balls?

Yes, the precise scale method can be applied to any number of balls. However, the number of balls must be a multiple of 3 in order for the method to work effectively.

## 4. Are there any limitations to the precise scale method?

One limitation of the precise scale method is that it can only be used for balls that are identical in weight. It also requires a very precise scale in order to get accurate results.

## 5. How does the precise scale method compare to other methods of solving the weighing balls problem?

The precise scale method is considered to be one of the most efficient and accurate methods for solving the weighing balls problem. It requires only one weighing and can be used for any number of balls, making it a preferred method for many scientists and mathematicians.

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