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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I have been insisting to my statistics students that for probabilities, the rule is the number of significant figures is the number of digits past the leading zeros or leading nines. For example to give 4 significant figures for a probability: 0.000001234 and 0.99999991234 are the correct number of decimal places. That way the complementary probability can also be given to the same significant figures ( 0.999998766 and 0.00000008766 respectively). More generally if you have a value that...

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