## 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:

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

math grad - Interest: Number theory
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### 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.

• benvitalis says:

Same here. I couldn’t find either.

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

• benvitalis says:

Have you tried a sum of the form a! ± b! ± c! ± d! ?

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

• benvitalis says:

Very nice! Thanks for sharing this