Question and Answers Forum

All Questions      Topic List

Logic Questions

Previous in All Question      Next in All Question      

Previous in Logic      Next in Logic      

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

1, 3, 9, 27?

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

Commented by Matica last updated on 27/Dec/22

What if you divide into 5 pieces?

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

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)

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com