Solve for $\overline{xyz}: \overline{xyz}\times \overline{zyx}=\overline{xzyyx}$

[Albert1]

$\overline{xyz}$ is a three digits number
$\overline{zyx}$ is also a three digits number
if :$\overline{xyz}\times \overline{zyx}
=\overline{xzyyx}$
please find:$\overline{xyz}$

 

[Albert1]

$\overline{xyz}$ is a three digits number
$\overline{zyx}$ is also a three digits number
if :$\overline{xyz}\times \overline{zyx}
=\overline{xzyyx}$
please find:$\overline{xyz}$

Spoiler

$(100x+10y+z)(100z+10y+x)
=10000x+1000z+100y+10y+x-(1)$
for $10000xz<100000$
we have $xz<10$
compare both sides of(1):$xz=x\,\ ,or\,\, z=1$
(1) becomes :
$10000x+1000y(x+1)+100(x^2+y^2+1)+10y(x+1)+x=10000x+1000+100y+10y+x$
or $100y(x+1)+10(x^2+y^2+1)+y(x+1)=100+10y+y-(2)$
compare both sides of (2):
$\therefore y(x+1)<10,\,, and ,\, y(x+1)=y$
$for (x+1)>1 \,\,\therefore y=0 ,and \,\,x=3$
we get :$\overline {xyz}=301$