An integer N consisting of five distinct nonzero digits has the curious property that it is equal to the sum of all the different three-digit integers formed by the three digit permutations of its five digits. Find N.

(In reply to Solution by e.g.)

The way the solution can be achieved is to realize that in the sum of the permutations of three digits out of the five, each of the digits appears 12 times in each of the three positions (hundreds, tens, units).  The reason for that is that there are 4 choices for the first of the other two positions and 3 choices for the second.  As a result of its appearing 12 times in each of these positions, the total is 12*111*(a+b+c+d+e) where the letters are the digits making up the 5-digit number, so

1332 (a+b+c+d+e) = 10000 a + 1000 b + 100 c + 10 d + e

Then a+b+c+d+e must be between 8 and 75 to produce a 5-digit number. Test each of these multiplied by 1332 to see if the resulting number's digits add to that number.

Posted by Charlie on 2004-06-30 09:42:42