Sorry proof only is that a=1, b could still be 2..9 but 913*923 933 943 953 963 973 983 993 842699 851829 860959 870089 879219 888349 897479 906609 the only palindrome is the last.
revised: >We know (from our brutish numerical experiments) that the answer is 913*993. >Why does that HAVE to be? >Well we know the answer must be of the form 9a3*9b3 because the first digit >must be maximal and so the last digit must also be a 9. Now the carry to the first digit must be exactly 1 (not 0 or >1) Ie 90000 <= (10000*9*(a+b))+(1000*a*b)+9 190000 > (10000*9*(a+b))+(1000*a*b)+9 assume a<=b let b be maximal (9) 90000 <= (10000*9*(a+9))+(1000*a*9)+9 10 000 <= (10000*(a+9))+(1000*a)+1 so a must be at least 1. assume 2 for a and b minimal, ie 2. Then 190000 > (10000*9*(2+2))+(1000*2*2)+9 190000 > (360000)+(4000)+9 false, so a=1. Assume b=8 90000 <= (10000*9*(9))+(1000*8)+9 10000 <= 90000+(9%~1000*8)+1 -79999 <= 8000%9 Not enough, so a=1 and b=9 ie 913*993 ie 913*993 greg ~krsnadas.org ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
