Find the unique three-digit number that is equal to the sum of the factorials of its digits.
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Given ABC = A! + B! + C!
Now, 1! = 1
2! = 2
3! = 6
4! = 24
5! = 120
6! = 720
7! = 5040
ABC is a three-digit number. Hence, 7 can not be one of the digits as 7! is a four digit number. So, 7 is ruled out.
Using 1,2,3,4,5,6 the highest three digit number whih can be formed will always be less than 700 but 6! is greater than 700. Hence, 6 is also ruled out as one of the digits.
So, Now we are left with 1,2,3,4 and 5.
Now, As ABC is a three-digit number and ABC = A! + B! + C!, so one of the numbers has to be 5.
Now, even if we sum 1! + 2! + 3! + 4! + 5! = 153 which means hundred place digit will be 1 only.
Now, 5 can be at the unit place or at hundred’s place.
15X or 15X where X will be one of 2,3 and 4.
When , 15X = 1! + 5! + X! = 1+120 + X! = 121 + X! No value of X satisfies Here.
When 1X5 = 1! + X! + 5! = 1+X!+120 = 121 + X!, only X = 4 satisfies here.
So, ABC = 145