World!Of
Numbers

WON plate
115 |

[ October 9, 2001 ]
How many palindromes can you find
that are the products of two pandigital numbers ?

 Two pandigital numbers1023687954 and 2901673548which on their own defy all palindromicy(any rearrangements of their digits never produces palindromes)yet when multiplied together produce surprisingly this palindrome2970408257528040792

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 29-digit palindrome which is the product
of three 10-digit 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 !

A000115 Prime Curios! Prime Puzzle
Wikipedia 115 Le nombre 115
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