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Question-157332

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Question Number 157332 by cortano last updated on 22/Oct/21
Answered by Rasheed.Sindhi last updated on 22/Oct/21
6x−y=7m⇒y=6x−7m P(x)=54x^2 +3xy−2y^2 +z+2021 =54x^2 +3x(6x−7m)−2(6x−7m)^2 +z+2021 =54x^2 +18x^2 −21mx−2(36x^2 −84mx+49m^2 )+z+2021 =54x^2 +18x^2 −21mx−72x^2 +168mx−98m^2 +z+2021 =147mx−98m^2 +z+2021 =49×3mx−49×2m^2 + 49×41+12 +z =49(3mx−2m^2 +41)+12+z 49 ∣ p(x) ⇒49∣(12+z) ⇒z+12≡0(mod 49) z≡−12+49(mod 49 z≡37(mod 49) z=37+49k Minimum positive of z=37+49×0 z=37
$$\mathrm{6}{x}−{y}=\mathrm{7}{m}\Rightarrow{y}=\mathrm{6}{x}−\mathrm{7}{m} \\ $$$${P}\left({x}\right)=\mathrm{54}{x}^{\mathrm{2}} +\mathrm{3}{xy}−\mathrm{2}{y}^{\mathrm{2}} +{z}+\mathrm{2021} \\ $$$$=\mathrm{54}{x}^{\mathrm{2}} +\mathrm{3}{x}\left(\mathrm{6}{x}−\mathrm{7}{m}\right)−\mathrm{2}\left(\mathrm{6}{x}−\mathrm{7}{m}\right)^{\mathrm{2}} +{z}+\mathrm{2021} \\ $$$$=\mathrm{54}{x}^{\mathrm{2}} +\mathrm{18}{x}^{\mathrm{2}} −\mathrm{21}{mx}−\mathrm{2}\left(\mathrm{36}{x}^{\mathrm{2}} −\mathrm{84}{mx}+\mathrm{49}{m}^{\mathrm{2}} \right)+{z}+\mathrm{2021} \\ $$$$=\mathrm{54}{x}^{\mathrm{2}} +\mathrm{18}{x}^{\mathrm{2}} −\mathrm{21}{mx}−\mathrm{72}{x}^{\mathrm{2}} +\mathrm{168}{mx}−\mathrm{98}{m}^{\mathrm{2}} +{z}+\mathrm{2021} \\ $$$$=\mathrm{147}{mx}−\mathrm{98}{m}^{\mathrm{2}} +{z}+\mathrm{2021} \\ $$$$=\mathrm{49}×\mathrm{3}{mx}−\mathrm{49}×\mathrm{2}{m}^{\mathrm{2}} +\:\mathrm{49}×\mathrm{41}+\mathrm{12}\:+{z} \\ $$$$=\mathrm{49}\left(\mathrm{3}{mx}−\mathrm{2}{m}^{\mathrm{2}} +\mathrm{41}\right)+\mathrm{12}+{z} \\ $$$$\mathrm{49}\:\mid\:{p}\left({x}\right) \\ $$$$\Rightarrow\mathrm{49}\mid\left(\mathrm{12}+{z}\right) \\ $$$$\Rightarrow{z}+\mathrm{12}\equiv\mathrm{0}\left({mod}\:\mathrm{49}\right) \\ $$$$\:\:\:\:{z}\equiv−\mathrm{12}+\mathrm{49}\left({mod}\:\mathrm{49}\right. \\ $$$$\:\:\:{z}\equiv\mathrm{37}\left({mod}\:\mathrm{49}\right) \\ $$$$\:\:\:{z}=\mathrm{37}+\mathrm{49}{k} \\ $$$${Minimum}\:{positive}\:{of}\:{z}=\mathrm{37}+\mathrm{49}×\mathrm{0} \\ $$$$\:\:\:\:{z}=\mathrm{37} \\ $$$$ \\ $$
Commented by cortano last updated on 22/Oct/21
thank you. it great way
$${thank}\:{you}.\:{it}\:{great}\:{way} \\ $$
Commented by mr W last updated on 22/Oct/21
i think with the conditions given in the question there is no requested solution possible. here just an example which also fulfills the conditions: x=(1/7) y=−((43)/7) z=16 < 37
$${i}\:{think}\:{with}\:{the}\:{conditions}\:{given}\:{in} \\ $$$${the}\:{question}\:{there}\:{is}\:{no}\:{requested} \\ $$$${solution}\:{possible}. \\ $$$${here}\:{just}\:{an}\:{example}\:{which}\:{also} \\ $$$${fulfills}\:{the}\:{conditions}: \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{7}} \\ $$$${y}=−\frac{\mathrm{43}}{\mathrm{7}} \\ $$$${z}=\mathrm{16}\:<\:\mathrm{37} \\ $$
Commented by Rasheed.Sindhi last updated on 22/Oct/21
You′re very right sir! ActuallyI assumed x x and y positive integers. And this I did unconsciously.Anyway my solution is only right for positive integers and according to the question you′re right.It seems that the condition of being integer of x and y is ignored mistakenly.
$$\mathrm{You}'\mathrm{re}\:\boldsymbol{\mathrm{very}}\:\boldsymbol{\mathrm{right}}\:\mathrm{sir}!\:\mathrm{ActuallyI}\:\mathrm{assumed}\:\mathrm{x} \\ $$$$\:\mathrm{x}\:\mathrm{and}\:\mathrm{y}\:\mathrm{positive}\:\mathrm{integers}.\:\mathrm{And}\:\mathrm{this} \\ $$$$\:\mathrm{I}\:\mathrm{did}\:\mathrm{unconsciously}.\mathrm{Anyway}\:\:\mathrm{my}\:\mathrm{solution} \\ $$$$\mathrm{is}\:\mathrm{only}\:\mathrm{right}\:\mathrm{for}\:\mathrm{positive}\:\mathrm{integers}\:\mathrm{and} \\ $$$$\mathrm{according}\:\mathrm{to}\:\mathrm{the}\:\mathrm{question}\:\mathrm{you}'\mathrm{re} \\ $$$$\mathrm{right}.\mathrm{It}\:\mathrm{seems}\:\mathrm{that}\:\mathrm{the}\:\mathrm{condition}\:\mathrm{of} \\ $$$$\mathrm{being}\:\mathrm{integer}\:\mathrm{of}\:\mathrm{x}\:\mathrm{and}\:\mathrm{y}\:\mathrm{is}\:\mathrm{ignored} \\ $$$$\mathrm{mistakenly}. \\ $$
Commented by mr W last updated on 22/Oct/21
i think so too. x,y should be integer.
$$\:{i}\:{think}\:{so}\:{too}.\:{x},{y}\:{should}\:{be}\:{integer}. \\ $$
Commented by Rasheed.Sindhi last updated on 22/Oct/21
Salute to your fine observation!
$$\mathrm{Salute}\:\mathrm{to}\:\mathrm{your}\:\mathrm{fine}\:\mathrm{observation}! \\ $$
Answered by ajfour last updated on 22/Oct/21
6x−y=7m ; m∈Z −z=54y^2 (t^2 +(t/(18))−(1/(27)))+p f=49N=54(x−((2y)/9))(x+(y/6))+z+p N=((54(x−((2y)/9))(x+(y/6))+(z+p))/7) 28N =(42m−2y)(7m+2y) +2(z+p) ⇒ z is a +min when 2(z+p)= 28N+(2y+7m)(2y−42m) is a +min. ⇒ 2z=−4042+28N−((63×35m^2 )/4) z =((7(16N−315m^2 ))/8)−2021 how to obtain a positive min out from this, i just dunnow..
$$\mathrm{6}{x}−{y}=\mathrm{7}{m}\:\:\:\:;\:\:{m}\in\mathbb{Z} \\ $$$$−{z}=\mathrm{54}{y}^{\mathrm{2}} \left({t}^{\mathrm{2}} +\frac{{t}}{\mathrm{18}}−\frac{\mathrm{1}}{\mathrm{27}}\right)+{p} \\ $$$${f}=\mathrm{49}{N}=\mathrm{54}\left({x}−\frac{\mathrm{2}{y}}{\mathrm{9}}\right)\left({x}+\frac{{y}}{\mathrm{6}}\right)+{z}+{p} \\ $$$${N}=\frac{\mathrm{54}\left({x}−\frac{\mathrm{2}{y}}{\mathrm{9}}\right)\left({x}+\frac{{y}}{\mathrm{6}}\right)+\left({z}+{p}\right)}{\mathrm{7}} \\ $$$$\:\:\mathrm{28}{N}\:\:=\left(\mathrm{42}{m}−\mathrm{2}{y}\right)\left(\mathrm{7}{m}+\mathrm{2}{y}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{2}\left({z}+{p}\right) \\ $$$$\Rightarrow\:\:\:{z}\:{is}\:{a}\:+{min}\:\:{when}\:\:\mathrm{2}\left({z}+{p}\right)= \\ $$$$\:\:\:\mathrm{28}{N}+\left(\mathrm{2}{y}+\mathrm{7}{m}\right)\left(\mathrm{2}{y}−\mathrm{42}{m}\right) \\ $$$$\:\:\:{is}\:{a}\:+{min}.\:\:\: \\ $$$$\Rightarrow\:\mathrm{2}{z}=−\mathrm{4042}+\mathrm{28}{N}−\frac{\mathrm{63}×\mathrm{35}{m}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\:\:{z}\:=\frac{\mathrm{7}\left(\mathrm{16}{N}−\mathrm{315}{m}^{\mathrm{2}} \right)}{\mathrm{8}}−\mathrm{2021} \\ $$$$\:\: \\ $$$${how}\:{to}\:{obtain}\:{a}\:{positive}\:{min} \\ $$$${out}\:{from}\:{this},\:{i}\:{just}\:{dunnow}.. \\ $$
Commented by cortano last updated on 22/Oct/21
thank you
$${thank}\:{you} \\ $$
Commented by mr W last updated on 22/Oct/21
ajfour sir is right. there is no unique solution.
$${ajfour}\:{sir}\:{is}\:{right}.\:{there}\:{is}\:{no}\:{unique}\: \\ $$$${solution}. \\ $$

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