I've attempted this problem by counting pages, which is a tedious approach, is there a shorter method?

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  • 3
    $\begingroup$ What have you tried? Surely you can do the first part, for example. $\endgroup$ – lulu Oct 9 '18 at 12:55
  • $\begingroup$ Hint, doubles go from 10 to 99. $\endgroup$ – Phil H Oct 9 '18 at 13:09

$$\begin{align}1\cdots9&\to1\cdots9, \\10\cdots99&\to11\cdots189, \\100\cdots999&\to192\cdots2889, \\1000\cdots9999&\to2893\cdots38889, \\&\cdots\end{align}$$





The limit digit counts are found from the summation



Formal count:

  • there are $1 \times 9=9$ digits from $1-9$

  • there are $2 \times 90=180$ digits from $10-99$

  • there are $3 \times 900=2700$ digits from $100-999$

But In third case we have more than the required one. So add first two to get $189$ digits and then subtract $189$ from $1260$ to see we have actually need $1071$ digits in third case. So divide $1071$ by $3$ to get $357$ numbers need from $100$ to $999$. So add $100$ to $356$ ( not add $357$ because we start from $100$) to get that number, namely $ 456$

So we have totally $456$ pages!


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