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Prove or disprove

A positive integer is divisible by 11 if and only if the sum of the two-digit blocks of its digits is divisible by 11

"like 10615=11.965 and 01+06+15=22 " I think it is true statement

I face difficulty to prove it for both side

n=$a_0$ +$a_1$ (100) +........+$a_k$ (100)$^k$

n≡ 0(mod 11)

any help with that please?

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  • $\begingroup$ Note that $100\equiv 1 (\bmod 11)$, therefore $a_1 \cdot 100 \equiv a_1 (\bmod 11);$ $a_2 \cdot 100^2 \equiv a_2 (\bmod 11)$; ... $\endgroup$ – Oleg567 Dec 15 '17 at 13:48
  • $\begingroup$ This isn't clear. If the number is, say, $121$, what sum do you take? $12+21$? Something else? $\endgroup$ – lulu Dec 15 '17 at 13:48
  • $\begingroup$ @lulu "like 10615=11.965 and 01+06+15=22 " $\endgroup$ – dr.rise Dec 15 '17 at 13:52
  • $\begingroup$ @Oleg567 "like 10615=11.965 and 01+06+15=22 " $\endgroup$ – dr.rise Dec 15 '17 at 13:53
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    $\begingroup$ You have over 300 reputation points, so there's no excuse for not using MathJax. $\endgroup$ – B. Goddard Dec 15 '17 at 14:21
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Hint: Note that $11\mid99$ and $100-1\mid100^k-1$.

The latter is because $$ \frac{x^k-1}{x-1}=x^{k-1}+x^{k-2}+\cdots+1 $$ Now write out the number as $$ n=\sum_{k=0}^n\overbrace{\ \ \ \ \ d_k\ \ \ \ \ }^{\text{$2$ digit blocks}}100^k$$ and the sum of the two digit blocks as $$ s=\sum_{k=0}^nd_k$$ then show $99\mid n-s$.

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  • $\begingroup$ how can I prove that please? $\endgroup$ – dr.rise Dec 15 '17 at 14:01
  • $\begingroup$ @dr.rise: does my edit answer your question? $\endgroup$ – robjohn Dec 15 '17 at 14:02
  • $\begingroup$ actually not yet , could you please explain it more ? $\endgroup$ – dr.rise Dec 15 '17 at 14:09
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The detailed proof is given in this pdf.
http://www.math.ucla.edu/~radko/circles/lib/data/Handout-530-641.pdf

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  • $\begingroup$ Please try to give a summary of any externally linked material in case that link should go stale. $\endgroup$ – robjohn Dec 15 '17 at 13:56
  • $\begingroup$ That article outlines another divisibility by $11$ method, not this one. It is possible to arrive at this one, but it is not immediate. $\endgroup$ – robjohn Dec 15 '17 at 13:59
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Consider any integer $n$.

Then we have:

$$n = a_0 + a_1 \cdot 100 + a_2 \cdot 100^2 + a_3 \cdot 100^3 + \cdots + a_{k-1} \cdot 100^{k-1} + a_k \cdot 100^k$$

where $0 \leq a_j \leq 99$ for each $0 \leq j \leq k.$

Consider $n$ modulo $11$:

$$n = a_0 + a_1 \cdot 1 + a_2 \cdot 1^2 + a_3 \cdot 1^3 + \cdots + a_{k-1} \cdot 1^{k-1} + a_k \cdot 1^k = a_0 + a_1 + a_2 + a_3 + \cdots + a_{k-1} + a_k \text{mod}(11).$$

Therefore, if $n = 0 \text{mod}(11)$ then $ a_0 + a_1 + a_2 + a_3 + \cdots + a_{k-1} + a_k = 0 \text{mod}(11)$ and vice versa (each form is equivalent when taken modulo 11) which is to say that $n$ is divisible by $11$ if and only if the sum of its two digit blocks is divisible by $11$.

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