Find the sum of digits of the product (16100) × (125135).

Concept: To find the digit sum of a huge product like this, rewrite every base as a prime power, combine matching exponents into a power of 10, and isolate the leftover coefficient — only that coefficient's digits matter, since the trailing zeros contributed by the power of 10 add nothing to the digit sum.

Application:

1. Write 16 as 24, so 16100 = 2400.

2. Write 125 as 53, so 125135 = 5405.

3. Combine: 2400 × 5405 = 2400 × 5400 × 55 = (2×5)400 × 55 = 10400 × 55.

4. Evaluate the leftover power: 55 = 3125.

5. So the product equals 3125 followed by 400 zeros — only the digits of 3125 affect the sum.

6. Sum of digits: 3 + 1 + 2 + 5 = 11.

Cross-check: Building 55 step by step — 51=5, 52=25, 53=125, 54=625, 55=3125 — confirms the leftover coefficient, and the digit sum 3+1+2+5 = 11 matches the option keyed as correct.