Find, analytically, all three digit numbers that equal the sum of the factorials of their digits. That is:
ABC = A! + B! + C!
(In reply to Answer
by K Sengupta)
Let us suppose that A >=7. if so, then the rhs is > 7! =5040, which is a four digit number. This is a contradiction. hence, A<=6.
If a=6, then rhs>= 6! = 720, which is a contradiction, since in that situation - the lhs <= 699. hence A<=5.
Let us now suppose that 3<= A <=5. obviously each of B and C must be less than 6, otherwise rhs >=720, which is a contradiction.
If A =3, then lhs >=300, but rhs <= 3!+5!+5! = 246, which is a contradiction.
If A =4, then lhs >=400, but rhs <= 4!+5!+5! = 264, which is a contradiction.
If A =5, then lhs >=500, but rhs <= 5!+5!+5! = 360, which is a contradiction.
Thus our supposition 3<= A <=5 is false, and accordingly A = 1, or 2.
SSuppose that A =2. Since the lhs consists of three digits and 2! =2, this is possible only when at least one of B and C is 5. However, we note that 2! + 5! = 122, and if the other value amongst B and C is <=4, then the rhs< = 122+24 = 146, while he lhs >= 200. accordingly, we must have B=C=5, so that:
lhs = 255 >= 242 = 2!+5!+5! = rhs, which leads to a contradiction.
Therefore, A must be equal to 1. In that situation, in terms of similar reasoning as above, at least one amongst b and C must be equal to 5.
If B=5, then:
150+C = 121+C!
-> C! - C = 29, which is not satisfied for any integer value of C.
If C=5, then:
105 + 10b = 121 + B!
-> 10B - B! = 16 .....(#)
Since B=5 would force a number greater than 200 in the lhs, we must have B<=4. Substituting B=1,2,3,4 in turn, we observe that only B=4 satisfies (#).
Consequently, (A,B,C) = (1,4,5) constitutes the only solution to the puzzle under reference.