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

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The problem involves finding a three-digit number $\overline{xyz}$ such that when multiplied by its reverse $\overline{zyx}$, the result equals the five-digit number $\overline{xzyyx}$. Both $\overline{xyz}$ and $\overline{zyx}$ are defined as three-digit integers. The equation can be expressed as $\overline{xyz} \times \overline{zyx} = \overline{xzyyx}$. The challenge lies in determining the specific values of $\overline{xyz}$ that satisfy this equation. Solving this will require exploring the properties of three-digit numbers and their reversals.
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$\overline{xyz}$ is a three digits number
$\overline{zyx}$ is also a three digits number
if :$\overline{xyz}\times \overline{zyx}
=\overline{xzyyx}$
please find:$\overline{xyz}$
 
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Albert said:
$\overline{xyz}$ is a three digits number
$\overline{zyx}$ is also a three digits number
if :$\overline{xyz}\times \overline{zyx}
=\overline{xzyyx}$
please find:$\overline{xyz}$
$(100x+10y+z)(100z+10y+x)
=10000x+1000z+100y+10y+x-(1)$
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 :$\overline {xyz}=301$
 
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