Find, analytically, all three digit numbers that equal the sum of the factorials of their digits. That is:

ABC = A! + B! + C!

As given that ABC is a 3-digit number, none of the digits may be greater than 5.

As 7! = 5040, 8! = 40320, 9! = 362880 each are greater than three digits, the sum of the digits of any number containing any of these digits would be greater than three digits in length. Thus we exclude any number that contains a 7, 8 or 9. This includes 6, as 6! = 720, and any number that includes a 6 must have a sum greater than 720, which must then have a factorial sum greater than three digits in length.

The only factorial of three digits in length is 5! = 120. The next smaller factorial is 4! = 24. The factorial sum of 444 is 72, which is less than three digits, thus the solution(s) must include at least one 5.

The factorial sum of the largest number of three digits that begins with a 5, with no digit greater than 5, i.e., 555, is 360. This value is smaller than 500, thus the first digit can not be a 5.

As at least one of the digits is a 5 and value of 5!, i.e., 120, contains no 5's, the number of the factorial sum of all three must contain a 5 in either the tens digit or ones digit. As it is, there are only three combination of digits that accomplish this, one with the 5 as the tens digit and two with the 5 as the ones digit:

5! + 4! + 3! = (120 + 24 + 6) = 150
5! + 4! + 1! = (120 + 24 + 1) = 145, and
5! + 4! + 0! = (120 + 24 + 1) = 145

As can be seen, 145 is the one and only solution.

Edited on September 13, 2008, 2:17 pm

Posted by Dej Mar on 2008-09-13 14:11:15