# Weighing with counterbalances

Polish mathematician Hugo Steinhaus (1887-1972) is well known for his popular books *Mathematical Kaleidoscope* and *100 Problems*. The following has been translated by my son David from a Russian book *Problems and Reflections* with articles and problems gleaned from Steinhaus's magazine *Matematyka*. Originally the problem below has been narrated by Dr. Sylvester Sharadek. The translation is somewhat frivolous in this respect.

### Counterbalances

Doctor David has two childhood friends, one of whom is the owner of a small convenience store, and the other owns a farm market. The owner of the small shop deals strictly with powdered goods, such as coffee, sugar, and flower. The owner of the farm market sells mainly such products as cucumbers, tomatoes, lemons, fish, etc. counted a few items at a time.

Doctor David had two sets of weights in the basement which he had not used in a long time. One set had four weights of 10, 30, 90, and 270 grams, while the other had five weights of 10, 20, 40, 80, and 160. One day, Dr. David invited his two friends over for dinner and promised to give them each a set of weights, but only if they would each choose the set of weights which was most convenient for his business.

Which set of weights did the owner of the convenience store choose, and which set of weights was taken by the owner of the farm market?

## Solution

First of all, let's note that the set of four weights (10, 30, 90, & 270 grams) allows one to weigh, within 10 grams, any weight between 10 and 400 grams:

10 = 10 20 = 30 - 10 30 = 30 40 = 10 + 30 50 = 90 - 30 - 10 60 = 90 - 30 70 = 90 + 10 - 30 80 = 90 - 10 90 = 90 100 = 90 + 10 110 = 90 + 30 - 10 120 = 90 + 30 130 = 90 + 30 + 10 140 = 270 - 90 - 30 - 10 150 = 270 - 90 - 30 160 = 270 + 10 - 90 - 30 170 = 270 - 90 - 10 180 = 270 - 90 190 = 270 + 10 - 90 200 = 270 + 30 - 90 - 10 210 = 270 + 30 - 90 220 = 270 + 30 + 10 - 90 230 = 270 - 30 - 10 240 = 270 - 30 250 = 270 + 10 - 30 260 = 270 - 10 270 = 270 280 = 270 + 10 290 = 270 + 30 - 10 300 = 270 + 30 310 = 270 + 30 + 10 320 = 270 + 90 - 30 - 10 330 = 270 + 90 - 30 340 = 270 + 90 + 10 - 30 350 = 270 + 90 - 10 360 = 270 + 90 370 = 270 + 90 + 10 380 = 270 + 90 + 30 - 10 390 = 270 + 90 + 30 400 = 270 + 90 + 30 + 10 |

The second set of five weights (10, 20, 40, 80, & 160g) allows one to weigh, within 10 grams, any weight between 10 and 310 grams:

10 = 10 20 = 20 30 = 20 + 10 40 = 40 50 = 40 + 10 60 = 40 + 20 70 = 40 + 20 + 10 80 = 80 90 = 80 + 10 100 = 80 + 20 110 = 80 + 20 + 10 120 = 80 + 40 130 = 80 + 40 + 10 140 = 80 + 40 + 20 150 = 80 + 40 + 20 + 10 160 = 160 170 = 160 + 10 180 = 160 + 20 190 = 160 + 20 + 10 200 = 160 + 40 210 = 160 + 40 + 10 220 = 160 + 40 + 20 230 = 160 + 40 + 20 + 10 240 = 160 + 80 250 = 160 + 80 + 10 260 = 160 + 80 + 20 270 = 160 + 80 + 20 + 10 280 = 160 + 80 + 40 290 = 160 + 80 + 40 + 10 300 = 160 + 80 + 40 + 20 310 = 160 + 80 + 40 + 20 + 10 |

The owner of the convenience store puts the necessary weight on one side of his balance, and puts on the other, for instance, a bag of sugar, after which he pours out or adds more sugar until both sides of his balance are level with one another The owner of this convenience store could just as easily use the first set of weights as the second, but the set of four weights would allow him to weigh up to 400g, whereas the set of five would only allow him to weigh up to 310g. So this person would likely choose the set of four.

For the owner of the farm market, the set of five weights is definitely more convenient, and here is why. The owner of the farm market puts a cucumber, for example, on one side of his balance, and then has to add a weight to the second side of the balance that is equal to the weight of the vegetable. Let's assume that the cucumber weighs 100g. The owner of the farm market will act as follows. On the second, yet empty side of the balance, he will put the heaviest weight, which, obviously, weighs more than the cucumber. Afterwards, he will put the weight of 80g on the side with the cucumber, and that side will weigh more than the side with the 100g weight. Then he will put the 40g weight on the opposite side (which will be lowered as a result) and, finally, the 20g weight on the side with the cucumber, after which the two sides of the balance will be level.

This method will allow the owner of the farm market to weigh, within 10g, any amount of vegetables between 10 and 310g. This same method however, when used with the set of 10, 30, 90, & 270g, will not allow him to weigh a cucumber at 100g with the same accuracy. The owner of the farm market could, at best, determine that the cucumber weighs less than 140g. This way, for the owner of the farm market, the set of four weights would be less convenient than the set of five.

### References

- H. Steinhaus,
*Mathematical Snapshots*, umpteen edition, Dover, 1999

### Weighing Coins, Balls, What Not ...

- The Oddball Problem, B. Bundy
- Weighing 12 coins, Dyson and Lyness' solution
- Weighing 12 coins, W. McWorter
- Thought Less Mathematics, D. Newman
- Weighing with counterbalances
- Odd Coin Problems, J. Wert
- Six Balls, Two Weighings
- 12 Coins in Verse
- Six Misnamed Coins, Two Weighings
- A Fake Among Eight Coins
- A Stack of Fake Coins
- Five Coins - One Good, One Bad
- With One Weighing

|Contact| |Front page| |Contents| |Up|

Copyright © 1996-2018 Alexander Bogomolny

66322010