Consider all nine-digit numbers consisting of one each of the digits 1 through 9.

What is the sum of these numbers?

 

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>

= ½ *(p+1)*p*((p-1)!)*(10^p-1)/9<o:p></o:p>

 = ½ *(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