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Question Number 151988 by mathdanisur last updated on 24/Aug/21

if  x;y;z  are natural numbers  such that  2x^x  + y^y  = 3z^z   then find  ((2021x + 2022y + 2023z)/(x + y + z)) = ?

$$\mathrm{if}\:\:\mathrm{x};\mathrm{y};\mathrm{z}\:\:\mathrm{are}\:\mathrm{natural}\:\mathrm{numbers} \\ $$$$\mathrm{such}\:\mathrm{that}\:\:\mathrm{2x}^{\boldsymbol{\mathrm{x}}} \:+\:\mathrm{y}^{\boldsymbol{\mathrm{y}}} \:=\:\mathrm{3z}^{\boldsymbol{\mathrm{z}}} \\ $$$$\mathrm{then}\:\mathrm{find}\:\:\frac{\mathrm{2021}\boldsymbol{\mathrm{x}}\:+\:\mathrm{2022}\boldsymbol{\mathrm{y}}\:+\:\mathrm{2023}\boldsymbol{\mathrm{z}}}{\mathrm{x}\:+\:\mathrm{y}\:+\:\mathrm{z}}\:=\:? \\ $$

Commented by mathdanisur last updated on 25/Aug/21

Thank you Ser

$${Thank}\:{you}\:{Ser} \\ $$

Commented by Rasheed.Sindhi last updated on 25/Aug/21

x=y=z: 2x^x +x^x =3x^x ⇒3x^x =^(✓) 3x^x      ((2021x + 2022y + 2023z)/(x + y + z))                     = ((2021x + 2022x + 2023x)/(x + x + x))                     = ((6066x )/(3 x ))=2022

$$\mathrm{x}=\mathrm{y}=\mathrm{z}:\:\mathrm{2x}^{\mathrm{x}} +\mathrm{x}^{\mathrm{x}} =\mathrm{3x}^{\mathrm{x}} \Rightarrow\mathrm{3x}^{\mathrm{x}} \overset{\checkmark} {=}\mathrm{3x}^{\mathrm{x}} \\ $$$$\:\:\:\frac{\mathrm{2021}\boldsymbol{\mathrm{x}}\:+\:\mathrm{2022}\boldsymbol{\mathrm{y}}\:+\:\mathrm{2023}\boldsymbol{\mathrm{z}}}{\mathrm{x}\:+\:\mathrm{y}\:+\:\mathrm{z}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{2021}\boldsymbol{\mathrm{x}}\:+\:\mathrm{2022x}\:+\:\mathrm{2023x}}{\mathrm{x}\:+\:\mathrm{x}\:+\:\mathrm{x}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{6066}\boldsymbol{\mathrm{x}}\:}{\mathrm{3}\:\mathrm{x}\:}=\mathrm{2022} \\ $$

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