With pen and paper

In Factorisation is the key, we assume that the number was of the form \( abccba \). Since it is the factor of two 3-digit number, we have :

\[ \begin{align} abccba &= (100a + 10b + c)(100d + 10e + f)\\ &= 10000ad + 1000(bd + ae) + 100(cd + be + af) + 10(ce + bf) + cf \end{align} \]

Assuming the first digit is 9, then \( cf \) must be equal to 9 as well.

The only ways to make the last digit nine are:

\[ 1 * 9\\ 3 * 3\\ 7 * 7 \]

Thus, both number must start with 9 and end with either 1, 3, 7 or 9. We also know that \( 100a + 10b + c \) or \( 100d + 10e + f \) is divisible by 11. The only numbers divisible by 11 and ending with 1, 3, 7 or 9 in the \( [900; 999] \) are :

\[ 913\\ 957\\ 979 \]

This give us:

\[ \text{a = 9}\\ \text{b = 1, 5 or 7}\\ \text{c = 3, 7 or 9}\\ \]

Resulting in the numbers:

\[ (900 + 10 + 3)(900 + 10e + 3) = 824439 + 9130x\\ (900 + 50 + 7)(900 + 10e + 7) = 867999 + 9570x\\ (900 + 70 + 9)(900 + 10e + 1) = 882079 + 9790x \]

The first number implies that \( e \) is equal to 9 because if \( e \) was equal to 8, then \(824439 * 9130 * 8 = 897479 \) would not start with 9. With \( e = 9 \) we have \( 913 * 993 \) which is the correct answer. Both \( (900 + 50 + 7)(900 + 10e + 7) \) and \( (900 + 70 + 9)(900 + 10e + 1) \) give smaller palindrome.