My calculator isn't good enough!

The given sum is $S=\displaystyle\sum_{k=1}^{2015}k\cdot 10^k$ .

Since $10\equiv -1\pmod{11}$ , we have,

$S=\sum_{k=1}^{2015} k\cdot 10^k\equiv \sum_{k=1}^{2015} k\cdot (-1)^k\pmod{11}\\ \implies S\equiv -1+2-3+4-5+6-\ldots +2014-2015\pmod{11}$

Group the odd and even terms together and we have,

$S\equiv \sum_{k=1}^{2015}k\cdot (-1)^k\equiv\left(\sum_{i=1}^{1007} 2i\right)-\left(\sum_{i=0}^{1007}(2i+1)\right)\\\implies S\equiv -(2\times 0+1)+\sum_{i=1}^{1007}(2i-2i-1)\\\implies S\equiv -1+\sum_{i=1}^{1007}(-1)\equiv -1-1007\equiv -1008\equiv\boxed{4}\pmod{11}$

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Here's a generalization. Consider the sum upto $n$ instead of $2015$ and denote that sum by $S_n$ where $n\in\Bbb{Z^+}$ . We have,

$S_n\equiv\left\lfloor (-1)^n\cdot\frac n2\right\rfloor\pmod{11}$

To obtain this result, it suffices to observe the following:

$S_n\equiv\begin{cases}\left(\sum\limits_{i=1}^{(n-1)/2}2i\right)-\left(\sum\limits_{i=0}^{(n-1)/2}(2i+1)\right)~\textrm{for odd }n\\\left(\sum\limits_{i=1}^{n/2}2i\right)-\left(\sum\limits_{i=1}^{n/2}(2i-1)\right)~\textrm{for even }n\end{cases}\pmod{11}$

Then, proceed similarly as done in original solution. Get results for both odd and even cases and merge both the results properly.

Each two terms combined together (first and second, third and fourth .... ) mod 11 is equal to one, we have 1007 pairs and we are left with (1007-2015) mod 11 which is equal to "-7" mod 11, which equals 4

$10^{2n+1} \equiv -1 \pmod {11}.\\ 10^{2n} \equiv+1 \pmod {11}.\\ So~odd~term~is ~ -1, ~~even~~ is~ +1.\\ \therefore~from ~1~to~2014~even~ terms ~will~ accumulate~+1007. \\But~last term ~is~2015~an~0dd~-tive~term. \\ So~ net~we~get~-2015+1007= -1008.\\ -1008+1100=92.~~92\equiv ~4~\mod{11} .\\ ANSWER=4.$

guessing method

Start by reducing this to a pattern. 10 divided by 11 has a remainder of 10. 210 divided by 11 has a remainder of 1. 3,210 divided by 11 has a remainder of 9. 43,210 divided by 11 has a remainder of 2. The pattern continues as 8,3,7,4,6,5,5,6,4,7,3,8,2,9,10,1. Separate of numbers of ñ from even and you get 10,9,8,7,5,4,3 ,2,1,0,10,9... For odd and 1,2,3,4,5,6,7,8,9,10,0,1,2... Our n is 2015 is odd so we use the odd pattern and voila we get 4!