Sum up numbers from 1 to 100000 after removing zeroes Ok so i have numbers from 1 to 100000.I need to remove the zeroes from all numbers and sum up numbers from 1 to 100000.So if the number is 405 consider it as 45 , if number if 20039 consider it 239.
Is there any mathematical way to solve this, i mean any formula or anything.
 A: Given the Mathematica flag, you might consider posting this type of question on mathematica.stackexchange.com.  Nonetheless,
deleteZeros[n_] := FromDigits[DeleteCases[IntegerDigits[n], 0]];
Total[deleteZeros /@ Range[100000]]

(* Out: 3466789696 *)

A: It's slightly more natural to consider the numbers from $1$ to $99999$, since they form complete sets of numbers with a certain number of digits; we can then add $1$ for $100000$ in the end.
There are $\displaystyle\binom5k$ ways to select $k$ of the $5$ digits to be non-zero, and they all lead to the same numbers when the zeros are removed. Every non-zero digit can take the values $1$ through $9$, whose sum is $9(9+1)/2=45$. The digits can be chosen independently, so the total for $k$ non-zero digits is $45$ for a given digit times $9^{k-1}$ options for the remaining digits times $\underbrace{1\ldots1}_k=(10^k-1)/9$ for the sum of the valencies of the digits, for a grand total of
$$
\begin{align}
\sum_{k=0}^5\binom5k45\cdot9^{k-1}\frac{10^k-1}9
&=
\frac59\left(\sum_{k=0}^5\binom5k90^k-\sum_{k=0}^5\binom5k9^k\right)
\\
&=
\frac59\left(91^5-10^5\right)
\\
&=3466789695\;.
\end{align}
$$
Add the $1$ for $100000$, and the result coincides with Mark's.
A: Look. You may ask how many numbers in 1-100000 end with 5 (the looks like xxxxx5 where $x \in \{0,...9\}$) let it be $a_{15}$ , then how many numbers looks like xxxx5x let it be $a_{25}$ and so on .
After that you should add all $a_{ij}$, were $i\in\{1,2,3,4,5\}$ and $j\in\{1,...9\}$.
