## March 17, 2012

### The 4 Weight Problem

A merchant had a 40 kg measuring weight that broke into four
pieces as the result of a fall. When the pieces were subsequently
weighed, it was found that the weight of each piece was a whole
number of kilograms and that the four pieces could be used to
weigh every integral weight between 1 and 40 kg. What were the
weight of the pieces.

Note:Weights may be places in either pan of the balance.

devendraiiit said...

1 3 9 27

Anonymous said...

Weights are 1 kg ,3kg,9 kg,27 kg

Similar to Binary Number System.

Anonymous said...

The question should be clarified by saying "every integral weight from 1 to 40 ...". Between implies that 1 and 40 are not included.

Twmkey said...

5 10 10 15 ?

Anonymous said...

Weights are 5 KG,15 KG,20 KG

Darby McDermott said...

Well in order for every integral weight to be measured, at least one of the weights must be 1 kg. Then the remaining pieces could be 4 kg, 5 kg, and 30 kg so that all the pieces add up to 40 kg and you can easily weight it in quantities of 5s and 1s.
Darby

Anonymous said...

4, 8, 12, 16

Phen Pete said...

Anonymous said...

If you can play around with the weights, then i think there are lots of solutions. Like if you can do 2 weights to distinguish the weight. With numbers 2, 6, 13, 19. You cannot get exactly 39 but you can get 40 and 38=13+19+6. So 39 can be obtained by 2 weights, it is lighter than 40 but heavier than 38. Is it valid?

Anonymous said...

Hello. And Bye. Thank you very much.

Anonymous said...

Hello. And Bye. Thank you very much.

Anonymous said...

simply dropping by to say hi

swarna said...

Hi, this problem follows 3 to the power 0, 3 to the power 1, 3 to the power 2 and 3 to the power 3. If the weight is 121 again the same pattern continues. Is there any reason for this. Why it has to be 3 to the power0,1,2,3 and so on for weighing on both the side. If the weighing on one side only it follows 2 to the power series. Is there an explanation for this.

Anonymous said...