Question Number 122653 by mathocean1 last updated on 18/Nov/20
$${we}\:{define}\:{in}\:{base}\:{x}\:{the}\: \\ $$$${number}\:{y}\:{and}\:{z}\:{by}: \\ $$$${y}=\overline {\mathrm{123}}\:^{{x}\:} {and}\:{z}=\overline {\mathrm{201}}\:^{{x}} \\ $$$$\left.\mathrm{1}\right)\:{without}\:{knowing}\:{x},\:{write} \\ $$$${the}\:{product}\:{x}×{y}×{z}\:{in}\:{function} \\ $$$${of}\:{x}. \\ $$$$\left.\mathrm{2}\right)\:{we}\:{suppose}\:{that}\:{x}+{y}+{z}=\mathrm{92} \\ $$$${find}\:{x};{y};{z}. \\ $$$$ \\ $$
Answered by mr W last updated on 18/Nov/20
$${y}={x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{3} \\ $$$${z}=\mathrm{2}{x}^{\mathrm{2}} +\mathrm{1} \\ $$$${xyz}={f}\left({x}\right)={x}\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{3}\right)\left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{1}\right) \\ $$$$ \\ $$$${x}+{y}+{z}={x}+{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{3}+\mathrm{2}{x}^{\mathrm{2}} +\mathrm{1}=\mathrm{3}{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{4}=\mathrm{92} \\ $$$$\mathrm{3}{x}^{\mathrm{2}} +\mathrm{3}{x}−\mathrm{88}=\mathrm{0}\:\Rightarrow{x}\notin\mathbb{N} \\ $$$$\Rightarrow{question}\:{wrong}! \\ $$
Commented by mathocean1 last updated on 18/Nov/20
$${okay}\:{sir} \\ $$