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If it takes 909 digits to number the pages of a book starting with page 1, how many pages are in the book?

Sagot :

jacob193

Answer:

There would be [tex]339[/tex] pages in this book.

Step-by-step explanation:

Important: there are [tex](x - y + 1)[/tex] pages between page [tex]y[/tex] and page [tex]x[/tex] ([tex]x > y[/tex].)

There are [tex](9 - 1 + 1) = 9[/tex] pages between page [tex]\verb!1![/tex] and [tex]\verb!9![/tex]. Each page takes [tex]1\![/tex] digit to number. That would be [tex]9 \times 1 = 9[/tex] digits.

There are [tex](99 - 10 + 1) = 89 + 1 = 90[/tex] pages between page [tex]\verb!10![/tex] and [tex]\verb!99![/tex]. Each page takes [tex]2[/tex] digits to number. That would be [tex]90 \times 2 = 180[/tex] digits.

So far, [tex]9 + 180 = 189[/tex] digits are used. There are [tex]909 - 189 = 720[/tex] more digits to consider.

Each page between [tex]\verb!100![/tex] and [tex]\verb!999![/tex] takes [tex]3[/tex] digits to number. Hence, [tex]720[/tex] digits would number [tex]720 / 3 = 240[/tex] pages in that range.

Important: let [tex]x[/tex] denote the number of the last page in this book. [tex](x - 100 + 1)[/tex] would denote the number of pages that take [tex]3[/tex] digits each to number.

Reasoning above shows there are [tex]240[/tex] of these pages in total. That is:

[tex](x - 100 + 1) = 240[/tex].

Hence:

[tex]x = 240 + 99[/tex].

[tex]x = 339[/tex].

In other words, the last page of this book is numbered [tex]\verb!339![/tex]. There would be [tex]339[/tex] pages in this book.

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