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

Concept: For numbers expressed as prime powers, (am)n = amn, and when two bases multiply to a round number such as 2 × 5 = 10, grouping (2a × 5a) = 10a turns a product into a power of ten times whatever factor is left over. Multiplying by 10a only appends a zeros to that leftover factor's digits, so the digit sum of the whole product equals the digit sum of the leftover factor alone.

1. Write 16 as a power of 2: 16 = 24, so 16100 = (24)100 = 2400.

2. Write 125 as a power of 5: 125 = 53, so 125135 = (53)135 = 5405.

3. Multiply the two results, pairing as many 5's with 2's as possible: 2400 × 5405 = 2400 × 5400 × 55 = (2 × 5)400 × 55 = 10400 × 55.

4. Evaluate the leftover factor: 55 = 54 × 5 = 625 × 5 = 3125.

5. So the product equals 3125 followed by 400 zeros; every digit besides that leading '3125' block is 0, so the digit sum of the whole number is just the digit sum of 3125: 3 + 1 + 2 + 5 = 11.

Cross-check: 55 = 3125 can be verified directly (52 = 25, 53 = 125, 54 = 625, 55 = 3125), and re-adding its digits (3 + 1 + 2 + 5) confirms 11 independently of the exponent manipulation above.