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Méziriac Bachet weights problem

Méziriac Bachet weights problem


This problem was raised by the French mathematician Claude Gaspard Bachet de Méziriac 400 years ago

A merchant had a weight of 40 kg that fell to the ground and broke into 4 unequal parts. He took these weights to a scale and found that each one coincidentally had a weight that was equal to an integer number of kilograms and observed that with these 4 weights he could weigh loads of objects whose weight was an entire number of kilograms between 1 and 40.

How many kilograms does each of the 4 weights weigh?

Solution

The four weights that were formed when dividing the initial weight are 1Kg, 3Kg, 9Kg and 27kg and with them we can weigh any amount between 1 and 40 kilos by distributing one or more weights on one or both sides of the scale. If we assume that we use the right dish to place the load despite the weight distribution would be as follows:

Left saucerRight saucerTotal heavy
-1 kg1 kg
1 kg3 kg2 kg
-3 kg3 kg
-1 Kg + 3 Kg4 kg
1 Kg + 3 Kg9 kg5 kg
3 kg9 kg6 kg
3 kg1 Kg + 9 Kg7 kg
1 kg9 kg8 kg
-9 kg9 kg
-1 Kg + 9 Kg10 kg
1 kg3 Kg + 9 Kg11 kg
-3 Kg + 9 Kg12 kg
-1 Kg + 3 Kg + 9 Kg13 kg
1 Kg + 3 Kg + 9 Kg27 kg14 kg
3 Kg + 9 Kg27 kg15 kg
3 Kg + 9 Kg1 Kg + 27 Kg16 kg
1 Kg + 9 Kg27 kg17 kg
9 kg27 kg18 kg
9 kg1 Kg + 27 Kg19 kg
1 Kg + 9 Kg3 Kg + 27 Kg20 kg
9 kg3 Kg + 27 Kg21 kg
9 kg1 Kg + 3 Kg + 27 Kg22 kg
1 Kg + 3 Kg27 kg23 kg
3 kg27 kg24 kg
3 kg1 Kg + 27 Kg25 kg
1 kg27 kg26 kg
-27 kg27 kg
-1 Kg + 27 Kg28 kg
1 kg3 Kg + 27 Kg29 kg
-3 Kg + 27 Kg30 kg
-1 Kg + 3 Kg + 27 Kg31 kg
1 Kg + 3 Kg9 Kg + 27 Kg32 kg
3 kg9 Kg + 27 Kg33 kg
3 kg1 Kg + 9 Kg + 27 Kg34 kg
1 kg9 Kg + 27 Kg35 kg
-9 Kg + 27 Kg36 kg
-1 Kg + 9 Kg + 27 Kg37 kg
1 kg3 Kg + 9 Kg + 27 Kg38 kg
-3 Kg + 9 Kg + 27 Kg39 kg
-1 Kg + 3 Kg + 9 Kg + 27 Kg40 kg

You will find more information about this problem on Wikipedia