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 . So far it was easy. Next, we know that 6 can also not be in the number since which means that but we already saw that that is impossible.
So we already know that the number can only possibly have as its digits. Next, we will observe that since, if it would be, then , but since we conclude that that is impossible. Using similar arguments, we can easily show that are impossible as well. So we know for sure that A=1.
Now, there are only two options for B,C such that . Can you see which ones they are, and why Opalg's answer is the only correct one?