Pandigital numbers expressed a sum of factorials

 
 
For example,

958371246 \; = \; 3! \; + \; 5! \; + \; 7! \; + \; 9! \; + \; 12! \; + \; 12!

six factorials are used.

 
 

Is this the least number of factorials that produce a 9-digit pandigital integer?

What is the least number of factorials that produce a 10-digit pandigital integer?

 
 

Paul found a 10-digit pandigital number:

6! \; + \; 7! \; + \; 9! \; + \; 12! \; + \; 12! \; + \; 13! \; = \; 7185392640

 
 
 
 
 
 
 
 
 
 

 
 
 
 

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About benvitalis

math grad - Interest: Number theory
This entry was posted in Number Puzzles and tagged , . Bookmark the permalink.

6 Responses to Pandigital numbers expressed a sum of factorials

  1. paul says:

    I can’t find any 5 factorials and this is the least number of factorials to make a 10 digit pan digital

    6! + 7! + 9! + 12! + 12! + 13! = 7185392640

    Paul.

  2. paul says:

    There is another version if you like, any number that contains all the digits, in this case 1 – 9 including multiples of digits is also considered a pan digital number, in which case we have

    1! + 9! + 14! + 14! = 174356945281

    just 4 factorials

  3. pipo says:

    That one works with some fantasy according to me:
    abs(12! – 13! – 5! – 3!) = 5748019326
    abs(11! + 10! – 13! – 6!) = 6183475920

    pipo

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