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 That's a lotta numbers! (Posted on 2006-04-14)
Consider all nine-digit numbers consisting of one each of the digits 1 through 9.

What is the sum of these numbers?

 See The Solution Submitted by Jer Rating: 3.2500 (4 votes)

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 The Solution | Comment 3 of 10 |

tomarken's solution is in conformity with the various steps corresponding to my method and the said method is furnished hereunder:
Consider all possible  9 digit numbers satisfying conditions of the problem.These numbers  are now arranged, WLOG, in ascending order of magnitude.
Then, the number of times any given digit  i , where  1< i < 9, is the first digit in the said arrangement is (9-1)! = 8! =40320.
It can easily be observed that the said rule applies to any given digit  i , where  1< i < 9, is the jth ( 1< j < 9)  digit in the said arrangement is (9-1)! = 8! =40320.
Accordingly, the required sum
= ( 1+2+-------+9) * 111111111 * 40320 =  201599999798400.

 Posted by K Sengupta on 2006-04-14 12:19:36

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