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?
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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.
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$\begingroup$ Why are you multiplying by 2015 and then 2016? Isn't the multiplication after that how any 2015 and 2016 digit numbers there are? $\endgroup$ Mar 11, 2017 at 1:52
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$\begingroup$ @GerardL. Do you first understand how many different $2015$ digit numbers there are? There are $9\cdot 10^{2014}$ different $2015$ digit numbers. If you take two $2015$ digit numbers and concatenate them, how many digits long will the result be? There will be $2\cdot 2015$ digits. If you take three $2015$ digit numbers and concatenate them, how many digits long will the result be? There will be $3\cdot 2015$ digits. If you take as many $2015$ digit numbers as exist and concatenate them together, how many digits long will it be? $(9\cdot 10^{2014})\cdot 2015$ digits long. $\endgroup$ Mar 11, 2017 at 2:13
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$\begingroup$ Similarly in the concatenation of all possible $2016$ digit long numbers there will be $9\cdot 10^{2015}\cdot 2016$ digits total. After that, we figure out how many of the digits must be removed because they were zeroes and we recognize that a tenth of the digits "in the second position" need to be removed and a tenth of the digits "in the third position" need to be removed, etc... giving us a total of $2015\cdot 9\cdot 10^{2015}\cdot \frac{1}{10}$ digits needing to be removed, giving us the final result. $\endgroup$ Mar 11, 2017 at 2:15