Question-183583

Question Number 183583 by Acem last updated on 27/Dec/22

Commented by Acem last updated on 27/Dec/22

A stone of 40 kg, i′ve cut it into 4 diferent parts, now i can use them for measure weights 1−40 Kg Sure the qusetion is weight of each piece? But by a logical solution not by experiment, please!

$${A}\:{stone}\:{of}\:\mathrm{40}\:{kg},\:{i}'{ve}\:{cut}\:{it}\:{into}\:\mathrm{4}\:{diferent}\:{parts}, \\ $$$$\:{now}\:{i}\:{can}\:{use}\:{them}\:{for}\:{measure}\:{weights}\:\mathrm{1}−\mathrm{40}\:{Kg}\: \\ $$$$\:{Sure}\:{the}\:{qusetion}\:{is}\:{weight}\:{of}\:{each}\:{piece}? \\ $$$$\:{But}\:{by}\:{a}\:{logical}\:{solution}\:{not}\:{by}\:{experiment},\:{please}! \\ $$$$ \\ $$

Answered by CrispyXYZ last updated on 27/Dec/22

$$\mathrm{1},\:\mathrm{3},\:\mathrm{9},\:\mathrm{27}? \\ $$

Commented by Matica last updated on 27/Dec/22

$${What}\:{if}\:{you}\:{divide}\:{into}\:\mathrm{5}\:{pieces}? \\ $$

Commented by CrispyXYZ last updated on 27/Dec/22

Each piece of stone can be placed on the right(+), left(−), or not(0). we can consider trinary(base 3) there are 4 digits in total 3^0 +3^1 +3^2 +3^3 =40 each piece=3^0 , 3^1 ,3^2 ,3^3

$$\mathrm{Each}\:\mathrm{piece}\:\mathrm{of}\:\mathrm{stone}\:\mathrm{can}\:\mathrm{be}\:\mathrm{placed}\:\mathrm{on}\: \\ $$$$\mathrm{the}\:\mathrm{right}\left(+\right),\:\mathrm{left}\left(−\right),\:\mathrm{or}\:\mathrm{not}\left(\mathrm{0}\right). \\ $$$$\mathrm{we}\:\mathrm{can}\:\mathrm{consider}\:\mathrm{trinary}\left(\mathrm{base}\:\mathrm{3}\right) \\ $$$$\mathrm{there}\:\mathrm{are}\:\mathrm{4}\:\mathrm{digits}\:\mathrm{in}\:\mathrm{total} \\ $$$$\mathrm{3}^{\mathrm{0}} +\mathrm{3}^{\mathrm{1}} +\mathrm{3}^{\mathrm{2}} +\mathrm{3}^{\mathrm{3}} =\mathrm{40} \\ $$$$\mathrm{each}\:\mathrm{piece}=\mathrm{3}^{\mathrm{0}} ,\:\mathrm{3}^{\mathrm{1}} ,\mathrm{3}^{\mathrm{2}} ,\mathrm{3}^{\mathrm{3}} \\ $$

Commented by Rasheed.Sindhi last updated on 27/Dec/22

With 5 such ′measures′ (1,3,9,27,81) you can weigh upto 121 kg.

$${With}\:\mathrm{5}\:{such}\:'{measures}'\:\left(\mathrm{1},\mathrm{3},\mathrm{9},\mathrm{27},\mathrm{81}\right) \\ $$$${you}\:{can}\:{weigh}\:{upto}\:\mathrm{121}\:{kg}. \\ $$

Commented by Rasheed.Sindhi last updated on 27/Dec/22

you can break ino 2 pieces of any one (except 1 of course) to make it ′5 peices′. For example : 1,3,4,5,27 but this will make your question somewhat inefficient. Because four peices are minimumly required and with it the question has unique solution. (You can′t do this by 3 peices of 40 kg)

$${you}\:{can}\:{break}\:{ino}\:\mathrm{2}\:{pieces}\:{of}\:{any}\:{one} \\ $$$$\left({except}\:\mathrm{1}\:{of}\:{course}\right)\:{to}\:{make}\:{it} \\ $$$$'\mathrm{5}\:{peices}'.\:{For}\:{example}\:: \\ $$$$\mathrm{1},\mathrm{3},\mathrm{4},\mathrm{5},\mathrm{27}\:{but}\:{this}\:{will}\:{make}\:{your} \\ $$$${question}\:{somewhat}\:{inefficient}. \\ $$$${Because}\:{four}\:{peices}\:{are} \\ $$$$\:{minimumly}\:{required}\:{and}\:{with}\:{it} \\ $$$${the}\:{question}\:{has}\:{unique}\:{solution}. \\ $$$$\left({You}\:{can}'{t}\:{do}\:{this}\:{by}\:\mathrm{3}\:{peices}\:{of}\:\mathrm{40}\:{kg}\right) \\ $$

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