Consider all nine-digit numbers consisting of one each of the digits 1 through 9.
What is the sum of these numbers?
(In reply to
Some Hints For Solving The Problem by K Sengupta)
OK, so there are 9! permutations of digits. In each place, 1/9 (40320) of them will be a 9, 1/9 will be an 8...etc.
The sum of the digits 1-9 is 45, so I'm guessing the sum we're looking for is:
4500000000
450000000
45000000
4500000
450000
45000
4500
450
+ 45
----------
4999999995
x 40320
----------
201599999798400
Did I screw that up?
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Posted by tomarken
on 2006-04-14 11:49:48 |
Is this it?
