11So?
I've done the same!
3 Answers
111st thing to note is a!+b!+c! has to be a 3-digit number....
If all a,b,c exceed 5, then
abc\equiv 0(mod10)\\ \implies c=0 - a contradiction!
Thus atleast 1 has to be less than 5, so let c<5.
Also a and b must exceed 4.
So a & b can only be 5 or 6.
Easy to see none satisfies the eqn, so no such nos!
62abc is the number 100a+10b+c :)
sorry not to have specified that