A number is written this way :
12345678910111213…….201520162017, what is the remainder when this number is divided by 11?
What I've thought of is discuss them in group.
So I got
1~9:odd digits sum:25, even digits sum:20
10~99: odd digits:45*9, even digits:45*10
100~999: observe that each hundred digit and tenth digit of each three digit number can be
eliminated.
Then I am stuck here. Cause there are thousands of four-digit numbers behind and they can not be eliminated.
Is there any other more convenient way to solve this question?
Any help is appreciated.
Best Answer
We have that $$\begin{align}N:=123456\dots 20162017&= 10^{6952}\sum_{k=1}^9 k\cdot 10^{9-k} +10^{6772} \sum_{k=10}^{99} k 10^{2(99-k)}\\ &\qquad+10^{4072} \sum_{k=100}^{999} k10^{3(999-k)}+ \sum_{k=1000}^{2017} k10^{4(2017-k)}. \end{align}$$ Now we evaluate $N$ modulo $11$, (and therefore $(10)^n\equiv (-1)^n \pmod{11}$), $$N\equiv -\sum_{k=1}^9 k\cdot (-1)^{k}+ \sum_{k=10}^{99} k -\sum_{k=100}^{999} k(-1)^{k}+\sum_{k=1000}^{2017} k.$$ Can you take it from here?