Difficult Word Problem

**MATNTRNG** · September 3rd 2010, 07:15 PM

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.

**Soroban** · September 4th 2010, 04:53 AM

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?

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

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.

**undefined** · September 4th 2010, 05:22 AM

Originally Posted by Soroban

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.

**MATNTRNG** · September 5th 2010, 08:05 AM

Wow this is amazing. Thank you for looking at this

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