Answer: D
100p+10q+r = p!+q!+r!
=> Now, 6!=720 and 7! =5040. If 7 is one of digits, then the sum of the factorials becomes four digits number or more. Hence the numbers 7,8,9 can be neglected.
Consider 6! =720. But 7 cannot be there in hundred's place. Hence, we can neglect 6 also.
Now, 5!=120, 4!=24, 3!=6, 2!=2 and 1!=1. To get a three digit number, 5 has to be present in the number. But 5 cannot be in hundreds place as then the number greater than 500 which cannot be obtained as the sum of factorial.
Also maximum possible number is 5!+4!+3! = 150, Also 'p' cannot be zero as it is a three digit number. hence p=1.
Then different possible cases are 154, 153, 152, 125, 135, 145
From this only 145 satisfies the condition
Thus, (4+5) = 9.