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I'm trying to prove that a number $a_n=a_0+a_1 \cdot 10^1+...+c_n\cdot 10^n$ is divisible by 11 iff $a_0-a_1+a_2-...+(a_n)\cdot(-1)^{n+1}$ is divisible by 11.

I began my proof and I found out that I first need to prove that $10^k+(-1)^{k+1}$ is divisible by 11 for all $k\in \mathbb{N} $.
I tried to prove this with induction:

for $k=1$ we get $t_1=10^1+(-1)^{1+1}=11$

now I assume $t_k=10^k+(-1)^{k+1}$ is divisible by 11.

need to prove that: $\\$ $t_{k+1}=10^{k+1}+(-1)^{k+2}=10^{k+1}+(-1)^{k}$ is divisible by 11.
I looked at:
$t_{k+1}-t_{k}=10^{k+1}+(-1)^{k}-10^k-(-1)^{k+1}=9\cdot 10^k +2\cdot (-1)^k$
and now I'm stuck :(

if you have any idea how can I complete my proof I will be very grateful!
Thank You All!

Arturo Magidin
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JollyQ1
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  • $10 = (11-1)$ so $10^k = (11-1)^k$. Applying binomial theorem finishes the problem. – JMoravitz Apr 21 '21 at 21:37
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    Consider instead $t_{k+1}-10t_k$ (which is to say, keep subtracting $t_k$ from that you have until interesting things happen). Much easier. – Arthur Apr 21 '21 at 21:38
  • Are you familiar with congruences or modular arithmetic, such as. $,10\equiv -1\pmod{11},$? – Bill Dubuque Apr 21 '21 at 22:05
  • $!\bmod 11!:\ 10\equiv -1\Rightarrow 10^k\equiv (-1)^k,$ by the Congruence Power Rule in the linked dupe (which is an inductive extension of the Product Rule). – Bill Dubuque Apr 21 '21 at 22:14
  • More generally casting out elevens is a special case of the Polynomial Congruence Rule, an inductive extension of both the Sum & Product Rules, see the 2nd linked dupe. – Bill Dubuque Apr 21 '21 at 22:21
  • In the 3rd linked dupe I explain how the inductive step (below) in a proof using divisibility language is just a special case of the Congruence Product Rule. $$\large \rm 11\mid \underbrace{10^n!-(-1)^n}{\Large a_n}\Rightarrow 11\mid 11(10^n)-(10^n!-(-1)^n)\ [= \underbrace{10^{n+1}!-(-1)^{n+1}}{\Large a_{n+1}}]\qquad $$ – Bill Dubuque Apr 21 '21 at 22:48
  • $10^k + (-1)^{k+1}= 11M$. so $10(10^k + (-1)^{k+1} = 10^{k+1} +(-1)^{k+1}10=110M$ So $10^{k+1} + (-1)^{k+1}10 - (-1)^{k+1}\cdot 11= 110M + - (-1)^{k+1}\cdot 11$. And so $10^{k+1} -(-1)^{k+1} = 10^{k+1} + (-1)^{k+2} = 11(10M - (-1)^{k+1})$. – fleablood Apr 22 '21 at 00:11

3 Answers3

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Note first that $$y^k-x^k = (y-x)\sum_{m=0}^{k} y^m x^{k-m}$$ (you can prove this for yourself as the right hand side is a telescoping sum). Then $$10^k + (-1)^{k+1} = 11 \sum_{m=0}^{k} (-1)^{k-m} 10^m$$ which is divisible by 11.

Adam B
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Hint $$ 10^{k+1}+(-1)^{k+2}=10^{k+1}+(-1)^{k+1}\cdot 10-(-1)^{k+1}\cdot 11=10 \cdot \left(10^{k}+(-1)^{k+1}\right)-(-1)^{k+1}\cdot 11 $$

Alternate alternate solution When $k=2l$ is even then $$ 10^{2l}-1=100^l-1=(100-1)( \mbox{ junk }) $$

When $k=2l+1$ is odd then $$ 10^{2l+1}+1=10^{2l+1}-10+11=10 (100^l-1) +11=10 (100-1)( \mbox{ junk }) +11 $$

N. S.
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This follows right away from the fact that $10\equiv-1\pmod{11}$. For then $10^n\equiv (-1)^n\pmod{11}$. So the alternating sum of the digits is congruent to the number modulo $11$.

J. W. Tanner
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