Let n = abc

where n = a × 100 + b × 10 + c a three digit number

a + b + c = S(n); Here note that S(n) ≤ 27

Since S(S(n)) = 2,

It means sum of digits of S(n) is 2

Now, S(n) can be, S(n) = 2, 11, 20 only

Now **Case-1 **

a + b + c = 2; possible cases are {0, 1, 1} gives 2 number and {2,0, 0} gives 1 number (ex : 200)

Total number in case (1) are 3

Total in case (3) = (3 × 4 + 6 × 4) = 36 ways

Total numbers = 3 + 61 + 36 = 100