Question

$1A\times A=9A$

Answer 1

For $A\times A=A$
From square tables $1,5,6$
$1\times 1=1$,
$5\times 5-25$,
$6\times 6=36$,
if $A=6$,
$16\times 6=96$
or
$(10+A)A=(90+A)$
$10A+A^2=90+A$
$A^2+9A-90=0$
$A^2+2.A\left(\frac{9}{2}\right)+{\left(\frac{9}{2}\right)}^2-{\left(\frac{9}{2}\right)}^2-90=0$
${(A+\frac{9}{2})}^2-\frac{81}{4}-90=0$
${(A+\frac{9}{2})}^2=\frac{441}{4}$
$A+\frac{9}{2}=\frac{21}{2}$
$A=\frac{12}{2}=6$