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That's a lotta numbers! (Posted on 2006-04-14) Difficulty: 2 of 5
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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Some Thoughts Generalised Result: Detailed Solution Not Given | Comment 4 of 10 |

 

In general, for any  p-digit number with digits from 1 to p;

The sum of all p -digit numbers consisting of one each of the digits 1 through p<o:p></o:p>

= (1+2+-------+p)*1111.11(p times)*(p-1)!<o:p></o:p>

= *(p+1)*p*((p-1)!)*(10^p-1)/9<o:p></o:p>

 = *(p+1)! *(10^p - 1)/9.<o:p></o:p>

 = (p+1)! *(10^p - 1)/18.<o:p></o:p>

 

 

Edited on April 15, 2006, 8:35 pm
  Posted by K Sengupta on 2006-04-14 12:28:12

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