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If you want to find b such that $a \equiv b \pmod{11}$, you do (assuming a has 4 digits, for example):

$$a_4*10^4+a_3*10^3+a_2*10^2+a_1*10^1+a_0*1$$

Then you calculate mod 11 for each product then add everything and do mod 11 again, like you do with other numbers.

The problem is that my theory book states that $10^k\equiv (-1)^k\pmod{11}$. I don't get this. How isn't the remainder of the division of 10 by 11 10? I thought that to calculate the remainder when your divisor is greater than the dividend, you to d*0 and your remainder is your dividend. For example, 10/11 = 0*11+10. Have I been doing it wrong?

Ethan
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SilenceOnTheWire
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4 Answers4

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"Mod" isn't an operation that you feed a number into and get a number out of, unless you're using computers. Mod makes an equivalence class; two numbers $x$ and $y$ are equivalent mod $m$ if $x = km + y$ for some integer $k$. This means you get a whole infinite set of numbers which are all equivalent mod $m$. For example, the equivalence class of 10 mod 11 is $$\{..., -12, -1, 10, 21, 32...\}$$ Frequently, the "useful" member of this equivalence class is the one you get out of the division theorem; the $r$ between 0 and your modulus. But these numbers are all equivalent by the mod relation, so they all share many arithmetic properties. In particular, if you take any $x$ and $y$ out of this set, and any positive integer $k$, then $x^k \equiv y^k \bmod m$. This is the property that the divisibility-by-11 rule uses.

Nicholas Viggiano
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  • No, the operator $\bmod$ is also used by mathematicians too (who also perform calculations, esp. (experimental) number theorists and algebraists). One shouldn't propagate ancient computational prejudices – Bill Dubuque Nov 18 '19 at 20:42
  • @BillDubuque No prejudice here, I'm a programmer by day. I thought the absolutism would help to provide clarity to a confused student rather than being exactly precise and potentially keeping them confused with "well, depending on context..." Though, I never claimed one could not be both mathematician and programmer, so I'm not sure your statement contradicts mine. – Nicholas Viggiano Nov 18 '19 at 20:54
  • Doing mathematical computation (such as using normal form operations) does not make one a "programmer". I know plenty of good mathematicians who do so but know very little about programming (and vice versa). It would be nice to get rid of these very old prejudices against calculation rather than continue to propagate them to future generations. Much of number theory and algebra was discovered by computation. – Bill Dubuque Nov 18 '19 at 21:00
  • @BillDubuque Again, I think you are reading malice or prejudice into my statements which does not exist there... I use "programmer" as a colloquial shorthand term and it is not meant to offend, snark, or divide. I know firsthand the usefulness of computation in these fields. I have edited my terminology to something perhaps less inciteful. – Nicholas Viggiano Nov 18 '19 at 21:15
  • Great, it's good to hear that no such prejudice was intended. Perhaps things aren't as bad as they were in the old days (e.g. when Artin used to "sneak" us requested Grobner computations to avoid embarassment). – Bill Dubuque Nov 18 '19 at 21:19
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While the actual remainder in dividing $10$ by $11$ is $10$ we subtract $11$ from it to make it $-1$

When we get to $100$ the remainder is $1$

So the remainder alternates between $1$ and $-1$

For example the remainder of $57$ is $7-5=2$

It is more fun working with larger numbers such as $5468$ whose remainder is $$8-6+4-5=1$$ Which is correct because $$5468=497(11)+1$$

Mohammad Riazi-Kermani
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Since $11\mid10-(-1)$, $$ 10\equiv-1\pmod{11} $$ That is, $a\equiv b\pmod{11}\iff11\mid a-b$.

Therefore, $$ \overbrace{\sum_{k=0}^nd_k10^k}^{\text{number}}\equiv\overbrace{\sum_{k=0}^nd_k(-1)^k}^{\substack{\text{alternating}\\\text{sum of digits}\\\text{of number}}}\pmod{11} $$ For example, $11\mid3583547$ since $$ \begin{align} &3+8+\,5+7=23\\ &3\,\,5\,\,8\,\,3\,\,5\,\,4\,\,7\\ &\,\,\,\ 5\,+3+4\,\,\,\,\,=\underline{12}\\ &\qquad\qquad\qquad\,\,11\leftarrow\text{difference is divisible by }11 \end{align} $$

robjohn
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  • $\displaystyle\left(358354 - 7 = 358347\right)\quad$$\displaystyle \left(35834 - 7 = 35827\right)\quad$$\displaystyle\left(3582 - 7 = 3575\right)\quad$$\displaystyle\left(357 - 5 = 352\right)\quad$$\displaystyle\left(35 - 2 = 33\right)\quad$$\displaystyle\left(3 - 3 = \color{red}{0}\right)$ – Felix Marin Jul 20 '20 at 20:16
  • So what you are saying is that $10a+b\equiv b-a\pmod{11}$, and, of course, $0\equiv-0\pmod{11}$. – robjohn Jul 20 '20 at 20:34
  • Sorry, I'm not $\displaystyle {\tt mod}$ fluenty. However, let $\displaystyle n \in \mathbb{N}{\ \ge\ 10}$ and $\displaystyle \ell{d}$ its last digit. Then, $$ n = 10\left(\color{red}{{n - \ell_{d} \over 10} - \ell_{d}}\right) + 11\ell_{d} $$ $\displaystyle 11 \mid \color{red}{red} \implies 11 \mid n$. Sorry again because it's a late comment. Thanks. – Felix Marin Aug 27 '20 at 00:12
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The important thing about the table below...

\begin{array}{ll} 10^1 = 11-1 & 10^2=99+1 \\ 10^3 = 1001-1 & 10^4 = 9999+1 \\ 10^5 = 100001-1 & 10^6 = 999999+1 \\ \qquad \vdots & \qquad \vdots \end{array}

... is that $11, 1001, 100001, \cdots$ and $99,9999,999999,\cdots$ are all multiples of $11$.

So, for example,

\begin{array}{ll} & gfedcba_{10} \\ = & 10^6g + 10^5f + 10^4e + 10^3d + 10^2c + 10b + a \\ = & (999999g+g)+(10001f - f)+(9999e + e)+(1001d - d)+(99c + c)+(11b - b)+(a) \\ = & (999999g+10001f+9999e+1001d+99c+11b+a) +(g-f+e-d+c-b+a) \\ = & (\text{something}\cdot 11)+(g-f+e-d+c-b+a) \end{array}

Steven Alexis Gregory
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