To: Mr. Patrick De Geest:
I note you are interested in palindromes.

Here is how I solve one of your problems analytically using sequence algebra:

```Problem stated:   312 x 221 = 68952
213 x 122 = 25986
-------------------------------------------------------------
Solution by Maple V R 3: (all lines below can be tested):
> A:=a(1)/x^1+a(2)/x^2+a(3)/x^3;
> B:=b(1)/x^1+b(2)/x^2+b(3)/x^3;
>AB:=expand(A*B);
>AA:=a(3)/x^1+a(2)/x^2+a(1)/x^3;
>BB:=b(3)/x^1+b(2)/x^2+b(1)/x^3;
>AABB:=expand(AA*BB);
>a(1):=3;
>a(2):=1;
>a(3):=2;
>b(1):=2;
>b(2):=2;
>b(3):=1;
-------------------------------------------------------------
Evaluation:  You will find:
>AB;
6/x^1+8/x^2+9/x^3+5/x^4+2/x^5
>AABB;
2/x^1+5/x^2+9/x^3+8/x^4+6/x^5
-------------------------------------------------------------
```
Further: So long as a(1) to a(3) and b(1) to b(3) are kept below 3, you should get all your reversed palindromes easily. Loss of palindromes will occur if any digits in AB or AABB exceed 10 due to carry overs operations. The limit seems to be 9 digits for the palindromes at which point all a(1) to a(5) and b(1) to b(5) must equal 1. There must be an analytical method from the above to handle other variations, e.g., to solve a set of linear equations in case digits exceed 10. Hope your readers will find an improvement. Huen Y.K. (CAHRC).
Dr Huen Y.K. (email) from CAHRC. SINGAPORE.