If f(x) = sum of all the digits of x, where x is a natural number, then what…

Solution:

Split the sum into two parts: numbers 101–199 and the single number 200.

• For 101–199 (99 numbers): each number has a hundreds digit 1, so the hundreds-place contribution is 99 × 1 = 99. The remaining contribution is the sum of digit-sums of the last two digits from 01 to 99.

• Sum of digit-sums for 00–99: tens digits sum to (0+1+…+9) × 10 = 45 × 10 = 450, and units digits sum to the same 450, so total = 900. (Since 00 contributes 0, this is also the sum for 01–99.)

• Therefore sum over 101–199 = 99 (hundreds contribution) + 900 = 999.

• Compute f(200) = 2 + 0 + 0 = 2.

• Add them: 999 + 2 = 1001.

Answer: 1001