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 Pandigital and divisible (Posted on 2010-12-26)
What percentage of pandigital* numbers is divisible by 990 ?

*A pandigital number consists of 10 distinct digits.

Try to solve analytically.

 See The Solution Submitted by Ady TZIDON No Rating

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 Correction of analytical solution Comment 4 of 4 |

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

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