Note that 40585 = 4! + 0! + 5! + 8! + 5! (i.e. the number can be expressed as the sum of the factorial of each of its digits).
There is only one three digit number that can be expressed this way. Find the sum of the digits of that number.
(P.S: I would really appreciate it if somebody could share their complete solutions to this problem.)
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Step 1) First of all please note that the required three digit number will never have the number >=6 in it as 6! = 720 you will see by trial and error that for this case you will get an exceeding number. Step 2) Now left are the numbers 5,4,3,2 and 1 . You will see in this case by placing 1! , 2! and 3! you will never get a 3 digit number. Step 3) Now simply manipulating the digit 4! and 5! in a set and with them joining 1!, 2!,3! in a set you will get 145 i.e 1!+4!+5!=145 whose sum of digits is 1+4+5=10