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From page 1 to page 9 we need 9 digits.

From page 10 to page 99 we need $(99-10+1) \times 2=180$ digits.

From page 100 to page 999 we need $(999-100+1) \times 3= 2700$ digits.

There are $2929-9-180-2700=40$ more to number four-number pages so $(a-1000+1)\times 4 = 40 (1)$ with $a$ is the final page's number.

Solve equation (1) for $a = 1009$ .

(I did many of these kinds of problem when I was in Grade 5)

Let the number of pages of a book be $p$ and the total number of digits of the page numbers be $d(p)$ . Then we note that:

$\begin{aligned} d(9) & = \underbrace 9_{\text{page 1 to 9}} \\ d(99) & = \underbrace 9_{1 \to 9} + \underbrace{2(99-9)}_{10 \to 99} = 189 \\ d(999) & = \underbrace{189}_{1 \to 99} + \underbrace{3(999-99)}_{100 \to 999} = 2889 \\ \implies d(p) & = \underbrace{2889}_{1 \to 999} + \underbrace{4(p-999)}_{1000 \to p} = 2929 \\ \implies 4(p-999) & = 2929 - 2889 \\ \implies p & = \frac {2929-2889}4 + 999 = \boxed{1009} \end{aligned}$