perplexus.info :: Probability : Pandigital and divisible

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.

Submitted by Ady TZIDON

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Solution: (Hide)

( 11* 5!* 4!)/(9!* 9) = 0.9700176% if no leading zeroes are allowed.
see Dan Rosen's posts.
Replace the 9*9! by 10! for the case of leading zeroes allowed.

Comments: ( You must be logged in to post comments.)

Subject	Author	Date
Correction of analytical solution	Dan Rosen	2011-04-01 23:20:55
Analytical Solution	Dan Rosen	2011-04-01 11:37:56
re: computer solution ahint for Ch	Ady TZIDON	2010-12-26 20:17:29
computer solution	Charlie	2010-12-26 19:45:19

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