# What is the easiest approch to solve this sequences and series problem?

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

• What have you tried? Surely you can do the first part, for example. – lulu Oct 9 '18 at 12:55
• Hint, doubles go from 10 to 99. – 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}

Hence

$$\color{blue}{11}=10+1\to11+1\times2=\color{green}{13}$$

and

$$\color{green}{456}=100+356\to192+356\times3=\color{blue}{1260}.$$

The limit digit counts are found from the summation

$$9+90\times2+900\times3+9000\times4+\cdots$$

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!