[ October 9, 2001 ]
How many palindromes can you find
that are the products of two pandigital numbers ?
Two pandigital numbers 1023687954 and 2901673548 which on their own defy all palindromicy (any rearrangements of their digits never produces palindromes) yet when multiplied together produce surprisingly this palindrome 2970408257528040792 .

There exists lots and lots more similar solutions.
Can you find them ? How many are there in total ?
[ Final score is 1277 ] _{January 2, 2009}
In collaboration with Peter Kogel.
I created a webpage twopan.htm where I display
all the palindromic products that we encountered.
Can you discover the first palindromes being the products
of three, four and more pandigitals ?
On [ April 20, 2012 ]
Carlos Rivera included the above 'three' question in his Puzzle 633
Soon after Giovanni Resta came with these smallest & largest solutions
1069458273 x 1082674593 x 1362840759
= 1577999653293663923569997751 [28]
9540138762 x 9568170243 x 9743625018
= 889414381197666666791183414988 [30]
Frank Rubin (email) wrote [ January 20, 2012 ]
I have found a 29digit palindrome which is the product
of three 10digit pandigital numbers.
2067945831 x 2758436091 x 3581704962
= 20431106772402320427760113402 [29]
This took me about 100 hours of computer time to find.
I estimate that there are somewhere between 80,000 and 320,000
solutions, so there's no way I'm going to try to find them all.
If there is some interest, I might try searching for
the smallest solution.
Peter Kogel (email) wrote [ October 28, 2008 ]
So far the closest I have come to success are the following :
38766662887833033878826666783 = 1264358097 x 8257640913 x 3713063103
37766638930788088703983666773 = 5769810243 x 8759416023 x 747259857
37766629851728882715892666773 = 2605894371 x 6291457803 x 2303563221
37766607369477677496370666773 = 1965024387 x 5610249387 x 3425767317
37766578522759495722587566773 = 1348296057 x 2659034781 x 10534122369

The problem is somewhat challenging to say the very least !
I began my search at a purely random point to test the efficiency
of my program and to try to gauge how long such a search might take.
Based on the progress I have made so far, I must confess that I'll
probably abandon the search, unless I can either think of a way
to radically improve the efficiency of my search algorithm or am
very lucky. I can see no reason why such a solution should not exist.
Perhaps one of your other readers can shed some light on the subject.
WONplate 97 investigated the ninedigital version of this topic !