Let n = abc
where n = a × 100 + b × 10 + c a three digit number
a + b + c = (n); Here note that (n) ≤ 27
Since ( (n)) = 2,
It means sum of digits of *(n) is 2
Now, (n) can be, (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
