1.19: Divisibility Tests for 7 and 13
- Page ID
- 83353
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)
( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\id}{\mathrm{id}}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\kernel}{\mathrm{null}\,}\)
\( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\)
\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\)
\( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)
\( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)
\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)
\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vectorC}[1]{\textbf{#1}} \)
\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)
\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)
\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Let \(a=a_ra_{r-1}\dotsm a_1a_0\) be the decimal representation of \(a\). Then
- \(7\mid a\Leftrightarrow 7\mid (a_r\dotsm a_1-2a_0)\).
- \(13\mid a\Leftrightarrow 13\mid (a_r\dotsm a_1-9a_0)\).
[Here \(a_r\dotsm a_1=\frac{a-a_0}{10}=a_r10^{r-1}+\dotsb+a_210+a_1\).]
Before proving this theorem we illustrate it with two examples. \[\begin{split} 7\mid 2481 &\Leftrightarrow 7\mid (248-2) \\ &\Leftrightarrow 7\mid 246 \\ &\Leftrightarrow 7\mid (24-12) \\ &\Leftrightarrow 7\mid 12 \end{split}\] since \(7\nmid 12\) we have \(7\nmid 2481\).
\[\begin{split} 13\mid 12987 &\Leftrightarrow 13\mid (1298-63) \\ &\Leftrightarrow 13\mid 1235 \\ &\Leftrightarrow 13\mid (123-45) \\ &\Leftrightarrow 13\mid 78 \end{split}\]
since \(6\cdot 13=78\), we have \(13\mid 78\). So, by Theorem \(\PageIndex{1}\) (b), \(13\mid 12987\).
Proof
of (a)
Let \(c=a_r\dotsm a_1\). So we have \(a=10c+a_0\). Hence \(-2a=-20c-2a_0\). Now \(1\equiv -20\pmod 7\) so we have \[-2a\equiv c-2a_0\pmod 7.\nonumber \] It follows from Theorem 1.17.1 that \[-2a\bmod 7=c-2a_0\bmod 7.\nonumber \] Hence, \(7\mid -2a\Leftrightarrow 7\mid (c-2a_0)\). Since \(\gcd(7,-2)=1\) we have \(7\mid -2a\Leftrightarrow 7\mid a\). Hence \(7\mid a\Leftrightarrow 7\mid (c-2a_0)\), which is what we wanted to prove.
of (b)
(This has a similar proof to that for \(\PageIndex{1}\)(a) and is left for the interested reader.)
Exercises
Using Theorem \(\PageIndex{1}\), determine which of the following, if any, is divisible by \(7\); also decide whether either is divible by 13. Show all your computations.
- \(6994\)
- \(6993\)
Give multiple reasons, including an application of Theorem \(\PageIndex{1}\), why no integer of the form \(10^n\) can be divisible by 7 or 13.
Using Theorem \(\PageIndex{1}\), determine the next year (after, possibly, the current year) that is
- divisible by 7.
- divisible by 13.
Show by example that, in the notation of Theorem \(\PageIndex{1}\), \(a\bmod 7\) need not be equal to \((a_r\dotsm a_1-2a_0) \bmod 7\).
Show that the number \(a = a_r a_{r-1} \cdots a_1 a_0\) is divisible by 13 if and only if \(a_r a_{r-1} \cdots a_1 + 4a_0\) is divisible by 13. (This gives a slightly different test for divisibility by 13.) Once you have proved this, apply it to test whether 111,111 is divisible by 13.