Which number has more digits and by how many?

Question

Paula writes down all 2015 digit positive integers into one row, forming a long number. Peter writes down all 2016 digit numbers into a long row and deleted all 0s. Whose number has more digits and by how many?
I tried to count how many 0s there are, but there are too many and it takes too long. What is the fast way to approach this?

Answer 1

There are $9\cdot 10^{2014}$ different $2015$ digit positive integers. By concatenating these together in some fashion, we get a total of $2015\cdot (9\cdot 10^{2014})$ digits in sequence.

There are $9\cdot 10^{2015}$ different $2016$ digit positive integers. A tenth of those will have a zero in the second digit. Similarly a tenth of them will have a zero in the third digit, fourth digit, etc...

Concatenating the integers together, we have $2016\cdot 9\cdot 10^{2015}$ however by deleting a zero from the second digit of each where it occurs will remove $\frac{1}{10}\cdot 9\cdot 10^{2015}$ digits. Similarly for each other digit position (except for the first of course since a $k$-digit number cannot start with zero). This brings our new count to $2016\cdot 9\cdot 10^{2015}-2015\cdot 9\cdot 10^{2015}\cdot \frac{1}{10} = (20160-2015)\cdot 9\cdot 10^{2014}$

You can immediately see that this number is larger than the first.