Hi, i am having trouble with this problem. Even a bit confused about the validity of the question, though....
A 3 digit no. ABC A>0 is such that A! + B! + C! becomes the no. itself. We have to find the value of B....
Obviously, 7,8 and 9 can not be in the number since $\displaystyle 9!>8!>7!>1000$. So far it was easy. Next, we know that 6 can also not be in the number since $\displaystyle 6!=720$ which means that $\displaystyle A\geq 7$ but we already saw that that is impossible.
So we already know that the number can only possibly have $\displaystyle \{0,1,2,3,4,5\}$ as its digits. Next, we will observe that $\displaystyle A \neq 5$ since, if it would be, then $\displaystyle B! + C! \geq 380 (= 500-120 = 500-5!)$, but since $\displaystyle B,C \in \{0,...,5\}$ we conclude that that is impossible. Using similar arguments, we can easily show that $\displaystyle A=4, A=3,A=2$ are impossible as well. So we know for sure that A=1.
Now, there are only two options for B,C such that $\displaystyle 100 \leq A!+B!+C! <200$. Can you see which ones they are, and why Opalg's answer is the only correct one?