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  1. #1
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    Difficult Word Problem

    In numbering the pages of a book, a printer used 3289 digits. How many pages were in the book, assuming that the first page in the book was number 1.
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  2. #2
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    Hello, MATNTRNG!

    There's no formula for this problem.
    I had to baby-talk my way through it.


    In numbering the pages of a book, a printer used 3289 digits.
    How many pages were in the book, assuming
    that the first page in the book was number 1?

    . . \begin{array}{cc}\text{Pages} & \text{Digits} \\ \hline<br />
1\text{ to } 9 & 9 \\<br />
10\text{ to }99 & 180 \\<br />
100\text{ to }999& 2700 \\ \hline<br />
\text{Total:} & 2889<br />
\end{array}


    Hence: . 3289 - 2889 \:=\:400 digits were used

    . . in the first 100 four-digit numbers.

    . . . . \{1000,\:1001,\:1002,\:\hdots\:1099\}


    Therefore, the book had 1099 pages.
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  3. #3
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    Quote Originally Posted by Soroban View Post

    There's no formula for this problem.
    I think you basically found a formula.

    \{a_n\} = 9, 180, 2700,\dots

    a_n=9\cdot10^{n-1}\cdot n

    \displaystyle b_n=\sum_1^n a_n

    Being lazy to derive, I looked up on OEIS (id:A033713 - OEIS Search Results)

    b_n=\frac{1}{9}\cdot(n\cdot(10^{n+1})-(n+1)\cdot10^n+1)

    So if the number of digits used is d, then we find the n such that

    b_n \le d < b_{n+1}

    and the number of pages is

    p=10^n-1+\frac{b_{n+1}-d}{n+1}

    But in fact the b_i are easily recognisable

    Using not-exactly-standard notation, b_n=[n-1](8\dots8)9 where the number of 8's is n-1.

    \{b_n\}=9,189,2889,38889,488889,\dots

    In other words, b_n = (n-1)\cdot10^n \ \ \ +\ \ \  (10^n-10)\cdot\frac{8}{9}\ \ \  +\ \ \ 9

    So there's no need to solve for n, we can tell just by looking.
    Last edited by undefined; September 4th 2010 at 05:37 AM.
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  4. #4
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    Wow this is amazing. Thank you for looking at this
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