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.

  • 3
    $\begingroup$ The portion of $4$-digit numbers is $2017+2016\cdot 10^4+...+1000\cdot 10^{4\cdot1017}$. The remainder of $10^4$ after division by $11$ is $1$. Therefore, this sum gives the same remainder as $2017+2016+...+1000=\frac{3017\cdot 1018}{2}$, which gives remainder $9$. You could do the other portions also this way, instead of the alternating sum of the digits. $\endgroup$
    – user752802
    Commented Feb 27, 2020 at 13:16

1 Answer 1


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?

  • $\begingroup$ Wouldn't the reversed number be $71026102...987654321$? $\endgroup$
    – supinf
    Commented Feb 27, 2020 at 14:20
  • $\begingroup$ @supinf Now it should be correct. Thanks for pointing out. $\endgroup$
    – Robert Z
    Commented Feb 27, 2020 at 15:43
  • $\begingroup$ @supunf What do you mean, please? $\endgroup$
    – Iris
    Commented Feb 27, 2020 at 23:32

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