A printer numbers the pages of a book starting with $1$ and uses $3189$ digits in all. How many pages does the book have?

## Concept & Logic This is the reverse of the standard page-numbering problem. We are given the total digits and must find the final page number. We do this by successively subtracting the known blocks of digits ($1$-digit, $2$-digit, $3$-digit numbers) until we find the range the final page falls into. ## Step-by-step Solution * **Determine standard digit blocks:** * Pages $1$ to $9$ use $9$ digits. * Pages $10$ to $99$ use $90 \times 2 = 180$ digits. * Pages $100$ to $999$ use $900 \times 3 = 2700$ digits. * Total digits for pages $1$ to $999 = 9 + 180 + 2700 = 2889$ digits. * **Calculate remaining digits:** * The total digits used is $3189$, which is greater than $2889$. This means the book has more than $999$ pages, so it has $4$-digit page numbers. * Remaining digits = $3189 - 2889 = 300$ digits. * **Find the number of 4-digit pages:** * Since each page from $1000$ onwards uses $4$ digits, we divide the remaining digits by $4$. * Number of $4$-digit pages = $\frac{300}{4} = 75$ pages. * **Calculate final page count:** * The book has $999$ pages (spanning $1-3$ digits) plus $75$ pages (spanning $4$ digits). $$ \text{Total Pages} = 999 + 75 = 1074 $$ ## Exam Strategy & Shortcut Memorizing the cumulative milestones is crucial: $1-99$ pages = $189$ digits $1-999$ pages = $2889$ digits Since $3189 > 2889$, you immediately know the answer is $999 + \frac{3189 - 2889}{4}$. $999 + \frac{300}{4} = 999 + 75 = 1074$. This takes less than $20$ seconds if you have $2889$ memorized. ## Common Pitfall A common error is finding that there are $75$ four-digit pages and then adding it to $1000$ instead of $999$, resulting in $1075$. Remember that the $75$ pages start *after* page $999$. ## Final Answer Therefore, the correct answer is **1074**.