Question 1116259
{{{99*99=9801}}} , so a product of two two-digit numbers is at most {{{9801}}} ,
and a palindrome that is a product of two two-digit numbers is less than {{{9801}}} .
The largest palindrome that is less than {{{9801}}} is {{{9779}}} .
 
If the first two digits are {{{a}}} and {{{b}}} in that order,
the number's digit read like {{{abba}}} ,
and the number is a multiple of {{{11}}} ,
because its value is
{{{1000a+100b+10b+a=1001a+110b=11*91a+11*10b=11(91a+10b)}}} .
If that {{{(91a+b)}}} factor can be further factored as
a two-digit number {{{t}}} times a single digit number {{{d}}} ,
the palindrome can be factored as the product of the two-digit numbers {{{11d}}} and {{{t}}} :
{{{11(91a+10b)=11*d*t=(11d)*t}}}

For example, {{{9779}}} , with {{{system(a=9,b=7)}}} can be factored as
{{{9779=11(91*9+10*7)=11(819+70)=11*889=11*7*127=77*127}}} ,
and does not work.
 
If we try all possible {{{b}}} values for {{{a=9}}} , we find
{{{matrix(2,8,
b,6,5,4,3,2,1,0,
91a+10b,879,869,859,839,829,819)}}} .
Out of all those values for {{{91a+b}}} ,
none is divisible by {{{2}}} , or {{{5}}} , or {{{7}}} ,
and only two of them are divisible by {{{3}}} .
That rules out all possible digit factors except {{{3}}} and {{{9}} .
Only one of those values, {{{819=9*91}}} , is divisible by {{{9}}} .
So, with {{{system(a=9,b=0,91a+10b=91*9=819)}}} ,
{{{highlight(9009)=11*819=11*9*91=99*91}}}
is a palindrome that can be written as a product of two two-digit numbers.
There will be others, such as {{{8118=11*738=11*9*82=99*82}}} ,
but {{{highlight(9009)}}} is the largest palindrome number that is the product of two 2-digit numbers.