Question 335299
Let the original number be ABC, where the numerical value is {{{100*A+10*B+C}}}.
The reverse number would be CBA or numerically {{{100*C+10*B+A}}}.
.
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{{{100A+10B+C-100C-10B-A=99A-99C=198}}}
{{{A-C=2}}}
{{{A=C+2}}}
A, B, and C can take on values from {{{0}}} to {{{9}}}.
For each value of B, there are {{{8}}} A,C values.
{{{0B2}}}
{{{1B3}}}
{{{2B4}}}
{{{3B5}}}
{{{4B6}}}
{{{5B7}}}
{{{6B8}}}
{{{7B9}}}
So there would be {{{10*8}}} or {{{80}}} number pairs that meet this criteria. 
If leading zeros aren't allowed (like 012/210), then reduce the B values allowed by 1, and then {{{9*8}}} or 72 number pairs are allowed.