Counting Digits Used in Page Numbering How many individual digits are required to number all pages of a book from page 1 through page 200 (inclusive)?

Introduction / Context:
Digit-counting problems are common in aptitude tests. The idea is to partition the page numbers by digit-length groups (1-digit, 2-digit, 3-digit) and sum their contributions.

Given Data / Assumptions:

• Pages run consecutively from 1 to 200 inclusive.
• Digits needed for a k-digit page number equals k.
• We compute by ranges: 1–9, 10–99, 100–200.

Concept / Approach:
Use place-value counting: the number of pages in each range times the digits per page in that range. Then add all contributions.

Step-by-Step Solution:
Pages 1–9: 9 pages * 1 digit = 9.Pages 10–99: 90 pages * 2 digits = 180.Pages 100–200: 101 pages * 3 digits = 303.Total digits = 9 + 180 + 303 = 492.

Verification / Alternative check:
Sanity check: A large portion comes from 3-digit pages (just over 300 digits), which aligns with the final total near 500.

Why Other Options Are Wrong:
200 and 372 underestimate by ignoring digit-length growth; 600 overshoots; 489 is a near-miss from miscounting pages 100–200.

Common Pitfalls:
Excluding page 200 or miscounting 100–199 as an even 100 pages (it is 101 pages from 100 through 200 inclusive).

Final Answer:
492