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Pandigital and divisible (Posted on 2010-12-26) |
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What percentage of pandigital* numbers is divisible by 990 ?
*A pandigital number consists of 10 distinct digits.
Try to solve analytically.
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Submitted by Ady TZIDON
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Solution:
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(Hide)
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( 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. |
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