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# 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 saucer Right saucer Total heavy - 1 kg 1 kg 1 kg 3 kg 2 kg - 3 kg 3 kg - 1 Kg + 3 Kg 4 kg 1 Kg + 3 Kg 9 kg 5 kg 3 kg 9 kg 6 kg 3 kg 1 Kg + 9 Kg 7 kg 1 kg 9 kg 8 kg - 9 kg 9 kg - 1 Kg + 9 Kg 10 kg 1 kg 3 Kg + 9 Kg 11 kg - 3 Kg + 9 Kg 12 kg - 1 Kg + 3 Kg + 9 Kg 13 kg 1 Kg + 3 Kg + 9 Kg 27 kg 14 kg 3 Kg + 9 Kg 27 kg 15 kg 3 Kg + 9 Kg 1 Kg + 27 Kg 16 kg 1 Kg + 9 Kg 27 kg 17 kg 9 kg 27 kg 18 kg 9 kg 1 Kg + 27 Kg 19 kg 1 Kg + 9 Kg 3 Kg + 27 Kg 20 kg 9 kg 3 Kg + 27 Kg 21 kg 9 kg 1 Kg + 3 Kg + 27 Kg 22 kg 1 Kg + 3 Kg 27 kg 23 kg 3 kg 27 kg 24 kg 3 kg 1 Kg + 27 Kg 25 kg 1 kg 27 kg 26 kg - 27 kg 27 kg - 1 Kg + 27 Kg 28 kg 1 kg 3 Kg + 27 Kg 29 kg - 3 Kg + 27 Kg 30 kg - 1 Kg + 3 Kg + 27 Kg 31 kg 1 Kg + 3 Kg 9 Kg + 27 Kg 32 kg 3 kg 9 Kg + 27 Kg 33 kg 3 kg 1 Kg + 9 Kg + 27 Kg 34 kg 1 kg 9 Kg + 27 Kg 35 kg - 9 Kg + 27 Kg 36 kg - 1 Kg + 9 Kg + 27 Kg 37 kg 1 kg 3 Kg + 9 Kg + 27 Kg 38 kg - 3 Kg + 9 Kg + 27 Kg 39 kg - 1 Kg + 3 Kg + 9 Kg + 27 Kg 40 kg