What percentage of pandigital* numbers is divisible by 990 ?


*A pandigital number consists of 10 distinct digits.

Try to solve analytically.

The division of the 9 digits 1...9 into 2 groups gives 11 (and not 6 ) possible groupings as follows :

      A                       B

   34678               1259

   34579               1268

   25678               1349

   24679               1358

   24589               1367

   23689               1457

   15679               2348

   14689               2357

   13789               2456

   12347              5689

   12356              4789

Therefore, the total different arrangements M will be :

        M= 5! * 4! * 11 = 31680

Additionaly - the total different pandigital perturbations excluding leading zeros should be :  9! * 9  and not 9!

Therefore the correct answer will be :

       ( 11* 5!* 4!)/(9!* 9) = 0.9700176%

 

 

Posted by Dan Rosen on 2011-04-01 23:20:55