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

ABC = A! + B! + C!

Since A!+B!+C! has to add up to a three digit number, it is clear none are 7, 8 or 9 (which would produce 4 or larger digit factorial). Thus, it follows none are 6 either (as 6!=720, so A=7, 8 or 9) However, using only numbers less than 5 would produce a one or two digit number (as 4!+4!+4! =72) So there must be at least one 5.

It is clear two or more digits can't be 5, as 240 + C! doesn't produce any fives for any C 1 to 5. Thus, the solution is 120!+_!+ _!.

If only one digit were 5, then it would need to be C. For A=5,  trying to get 380 or more with B and C less than 5 is impossible. For B=5, the only solution for A!+C! to be between 30 and 40 is 3!+4!=30, which doesn't work (150 has no 3 or 4).

Thus, C=5, and the only sum which ends in 5 is 1!+4!, which produces 145, the only solution

Posted by Gamer on 2008-09-13 16:26:18