MHB Finding the Frequency of 5 in $S$

[anemone]

Let $S=1+10+19+28+\cdots+10^{2013}$. How often does the digit 5 occur in $S$.

 

[chisigma]

Let $S=1+10+19+28+\cdots+10^{2013}$. How often does the digit 5 occur in $S$.

[sp]It is a series of arithmetic type ...

$\displaystyle S_{n} = \sum_{k=1}^{n} a_{k},\ a_{k} = a_{1} + d\ (k-1)\ (1)$

... where $d=9$, $a_{1}=1$ and...

$\displaystyle n= \frac{10^{m} - 1}{9} + 1\ (2)$

It is well known that...

$\displaystyle S_{n} = \frac{n}{2} (a_{1} + a_{n})\ (3)$

... and because $\frac{n}{2}$ is a number composed by m-2 digits 'five'and 1 digit 'sex' the numer of digit 'five' in $S_{n}$ is 2 (m-2). For m=2013 the number of digit 'five' is 2 (2013 - 2) = 4022...[/sp]

Kind regards

$\chi$ $\sigma$

 

[anemone]

[sp]It is a series of arithmetic type ...

$\displaystyle S_{n} = \sum_{k=1}^{n} a_{k},\ a_{k} = a_{1} + d\ (k-1)\ (1)$

... where $d=9$, $a_{1}=1$ and...

$\displaystyle n= \frac{10^{m} - 1}{9} + 1\ (2)$

It is well known that...

$\displaystyle S_{n} = \frac{n}{2} (a_{1} + a_{n})\ (3)$

... and because $\frac{n}{2}$ is a number composed by m-2 digits 'five'and 1 digit 'sex' the numer of digit 'five' in $S_{n}$ is 2 (m-2). For m=2013 the number of digit 'five' is 2 (2013 - 2) = 4022...[/sp]

Kind regards

$\chi$ $\sigma$

Hi chisigma,:)

This is one very easy to follow and nevertheless VERY elegant solution to a pretty hard challenge! How many thumbs up can I give to this solution?!?(Yes)(Yes)

Your reply kind of reassuring me to keep posting many more challenge problems here because it seems to me our members will just continue to surprise us by their wonderful and insightful solution!

Thanks chisigma for this solution! And thanks for participating!