{"id":759,"date":"2026-02-11T11:15:15","date_gmt":"2026-02-11T10:15:15","guid":{"rendered":"https:\/\/clases.jesussoto.es\/?p=759"},"modified":"2026-02-10T18:19:11","modified_gmt":"2026-02-10T17:19:11","slug":"el-algoritmo-de-la-division-con-maxima","status":"publish","type":"post","link":"https:\/\/clases.jesussoto.es\/?p=759","title":{"rendered":"El algoritmo de la divisi\u00f3n con maxima"},"content":{"rendered":"
\n

1. El algoritmo de la divisi\u00f3n<\/h2>\n<\/div>\n

<\/p>\n

Dados dos n\u00fameros enteros \\(a\\) y \\(b\\), con \\(a\\) no nulo, la divisi\u00f3n eucl\u00eddea asocia un cociente \\(q\\in\\mathbb{Z}\\) y un resto \\(r\\in\\mathbb{Z}\\), \u00fanicos, que verifican: \\[b=q\\,a+r,\\quad 0\\leq r<|a|\\]<\/div>\n

<\/p>\n\n\n
(%i6)<\/span><\/td>\nb<\/span>:<\/span>2<\/span>^<\/span>(<\/span>2<\/span>^<\/span>5<\/span>)<\/span>+<\/span>1<\/span>;<\/span>
<\/span>a<\/span>:<\/span>2<\/span>^<\/span>(<\/span>2<\/span>^<\/span>4<\/span>)<\/span>+<\/span>1<\/span>$<\/span>
<\/span>r<\/span>:<\/span>b<\/span>\u2212<\/span>a<\/span>$<\/span>
<\/span>i<\/span>:<\/span>1<\/span>$<\/span>
<\/span>while<\/span>(<\/span>r<\/span>><\/span>a<\/span>)<\/span> do <\/span>(<\/span>
<\/span> \u00a0\u00a0 r<\/span>:<\/span>r<\/span>\u2212<\/span>a<\/span>,<\/span>
<\/span> \u00a0\u00a0 i<\/span>:<\/span>i<\/span>+<\/span>1<\/span>
<\/span>)<\/span>$<\/span>
<\/span>print<\/span>(<\/span>\u00abcociente:\u00bb<\/span>,<\/span>i<\/span>,<\/span>\u00abresto\u00bb<\/span>,<\/span>r<\/span>,<\/span>\u00bb -> \u00ab<\/span>,<\/span>b<\/span>,<\/span>\u00ab=\u00bb<\/span>,<\/span>i<\/span>,<\/span>\u00ab*\u00bb<\/span>,<\/span>a<\/span>,<\/span>\u00ab+\u00bb<\/span>,<\/span>r<\/span>)<\/span>$<\/span><\/span><\/td>\n<\/tr>\n<\/table>\n

\\[4294967297\\]\\[\\mbox{cociente}:65535\\, \\mbox{resto}:2\\,\\to\\, 4294967297=65535\\cdot 65537+2\\]<\/p>\n

<\/p>\n

Veamos almacenando<\/div>\n

<\/p>\n\n\n
(%i13) <\/span><\/td>\nb<\/span>:<\/span>126<\/span>$<\/span>
<\/span>a<\/span>:<\/span>33<\/span>$<\/span>
<\/span>r<\/span>:<\/span>[<\/span>b<\/span>\u2212<\/span>a<\/span>]<\/span>$<\/span>
<\/span>i<\/span>:<\/span>1<\/span>$<\/span>
<\/span>while<\/span>(<\/span>r<\/span>[<\/span>i<\/span>]<\/span>><\/span>a<\/span>)<\/span> do <\/span>(<\/span>
<\/span> \u00a0\u00a0 r<\/span>:<\/span>append<\/span>(<\/span>r<\/span>,<\/span>[<\/span>r<\/span>[<\/span>i<\/span>]<\/span>\u2212<\/span>a<\/span>]<\/span>)<\/span>,<\/span>
<\/span> \u00a0\u00a0 i<\/span>:<\/span>i<\/span>+<\/span>1<\/span>
<\/span>)<\/span>$<\/span>
<\/span>print<\/span>(<\/span>r<\/span>)<\/span>$<\/span>
<\/span>print<\/span>(<\/span>\u00abcociente:\u00bb<\/span>,<\/span>i<\/span>,<\/span>\u00abresto\u00bb<\/span>,<\/span>r<\/span>[<\/span>i<\/span>]<\/span>,<\/span>\u00bb -> \u00ab<\/span>,<\/span>b<\/span>,<\/span>\u00ab=\u00bb<\/span>,<\/span>i<\/span>,<\/span>\u00ab*\u00bb<\/span>,<\/span>a<\/span>,<\/span>\u00ab+\u00bb<\/span>,<\/span>r<\/span>[<\/span>i<\/span>]<\/span>)<\/span>$<\/span><\/span><\/td>\n<\/tr>\n<\/table>\n

\\[\\left[ 93{,}60{,}27\\right]\\]\\[\\mbox{cociente}:3,\\,\\mbox{resto}:27\\,\\to\\, 126=3\\cdot 3+2\\]<\/p>\n

<\/p>\n

\n

1.1 cociente<\/h3>\n<\/div>\n

<\/p>\n

Por el algoritmo de la divis\u00f3n, el cociente de b\/a ser\u00e1 el entero q tal que
a*q<=b<a*(q+1)<\/div>\n

<\/p>\n

\n

1.1.1 algoritmo<\/h4>\n<\/div>\n

<\/p>\n\n\n
(%i15) <\/span><\/td>\ncociente<\/span>(<\/span>b<\/span>,<\/span>a<\/span>)<\/span>:<\/span>=<\/span>block<\/span>(<\/span>[<\/span>q<\/span>]<\/span>,<\/span>
<\/span> \u00a0\u00a0 q<\/span>:<\/span>1<\/span>,<\/span>
<\/span> \u00a0\u00a0 while<\/span>(<\/span>a<\/span>\u00b7<\/span>q<\/span><<\/span>=<\/span>b<\/span>)<\/span> do <\/span>(<\/span>
<\/span> \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 q<\/span>:<\/span>q<\/span>+<\/span>1<\/span>
<\/span> \u00a0\u00a0 )<\/span>,<\/span>
<\/span> \u00a0\u00a0 q<\/span>\u2212<\/span>1<\/span>
<\/span>)<\/span>$<\/span>
<\/span>cociente<\/span>(<\/span>126<\/span>,<\/span>33<\/span>)<\/span>;<\/span><\/span><\/td>\n<\/tr>\n<\/table>\n

\\[3\\]<\/p>\n

<\/p>\n

\n

1.1.2 Funci\u00f3n floor() de maxima<\/h4>\n<\/div>\n

<\/p>\n

floor(x) nos da el cociente:
Sea \\(\\lfloor\\cdot \\rfloor:\\mathbb{R}\\to\\mathbb{Z}\\), definida por,
\\[\\forall x\\in\\mathbb{R},\\ \\lfloor x\\rfloor=\\{n\\in\\mathbb{Z}:n\\leq x<n+1\\}\\]<\/div>\n

<\/p>\n\n\n
(%i16) <\/span><\/td>\nfloor<\/span>(<\/span>126<\/span>\/<\/span>33<\/span>)<\/span>;<\/span><\/span><\/td>\n<\/tr>\n<\/table>\n

\\[3\\]<\/p>\n

<\/p>\n

\n

1.2 resto<\/h3>\n<\/div>\n

<\/p>\n\n\n
(%i18) <\/span><\/td>\nresto<\/span>(<\/span>b<\/span>,<\/span>a<\/span>)<\/span>:<\/span>=<\/span>block<\/span>(<\/span>[<\/span>q<\/span>]<\/span>,<\/span>
<\/span> \u00a0\u00a0 q<\/span>:<\/span>1<\/span>,<\/span>
<\/span> \u00a0\u00a0 while<\/span>(<\/span>a<\/span>\u00b7<\/span>q<\/span><<\/span>=<\/span>b<\/span>)<\/span> do <\/span>(<\/span>
<\/span> \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 q<\/span>:<\/span>q<\/span>+<\/span>1<\/span>
<\/span> \u00a0\u00a0 )<\/span>,<\/span>
<\/span> \u00a0\u00a0 b<\/span>\u2212<\/span>a<\/span>\u00b7<\/span>(<\/span>q<\/span>\u2212<\/span>1<\/span>)<\/span>
<\/span>)<\/span>$<\/span>
<\/span>resto<\/span>(<\/span>126<\/span>,<\/span>33<\/span>)<\/span>;<\/span><\/span><\/td>\n<\/tr>\n<\/table>\n

\\[27\\]<\/p>\n

<\/p>\n

\n

1.2.1 Funci\u00f3n mod() de maxima<\/h4>\n<\/div>\n

<\/p>\n

la funci\u00f3n mod() nos da el resto:<\/div>\n

<\/p>\n\n\n
(%i19) <\/span><\/td>\nmod<\/span>(<\/span>126<\/span>,<\/span>33<\/span>)<\/span>;<\/span><\/span><\/td>\n<\/tr>\n<\/table>\n

\\[27\\]<\/p>\n

<\/p>\n

\n

1.3 Ejercicio<\/h3>\n<\/div>\n

<\/p>\n

\u00bfCu\u00e1l es el resto de la divisi\u00f3n de \\(F_5=2^{2^5}+1\\) por 13?<\/div>\n

<\/p>\n\n\n
(%i21) <\/span><\/td>\nF5<\/span>:<\/span>2<\/span>^<\/span>(<\/span>2<\/span>^<\/span>5<\/span>)<\/span>+<\/span>1<\/span>$<\/span>
<\/span>print<\/span>(<\/span>F5<\/span>,<\/span>\u00ab=\u00bb<\/span>,<\/span>floor<\/span>(<\/span>F5<\/span>\/<\/span>13<\/span>)<\/span>,<\/span>\u00ab*\u00bb<\/span>,<\/span>13<\/span>,<\/span>\u00ab+\u00bb<\/span>,<\/span>mod<\/span>(<\/span>F5<\/span>,<\/span>13<\/span>)<\/span>)<\/span>$<\/span><\/span><\/td>\n<\/tr>\n<\/table>\n

\\[\\]\\[4294967297=330382099{\\cdot}13+10\\]<\/p>\n

<\/p>\n

\n

1.4 Ejercicio<\/h3>\n<\/div>\n

<\/p>\n

\u00bfCu\u00e1ntos n\u00fameros hay entre 123 y 235 divisibles por 7?<\/div>\n

<\/p>\n\n\n
(%i25) <\/span><\/td>\na<\/span>:<\/span>[<\/span>]<\/span>$<\/span>
<\/span>for <\/span>i<\/span>:<\/span>126<\/span> thru <\/span>235<\/span> do <\/span>(<\/span>
<\/span> \u00a0\u00a0 if<\/span>(<\/span>mod<\/span>(<\/span>i<\/span>,<\/span>7<\/span>)<\/span>=<\/span>0<\/span>)<\/span> then <\/span> \u00a0
<\/span> \u00a0\u00a0\u00a0\u00a0\u00a0 a<\/span>:<\/span>append<\/span>(<\/span>a<\/span>,<\/span>[<\/span>i<\/span>]<\/span>)<\/span> \u00a0
<\/span>)<\/span>$<\/span>
<\/span>a<\/span>;<\/span>
<\/span>length<\/span>(<\/span>a<\/span>)<\/span>;<\/span><\/span><\/td>\n<\/tr>\n<\/table>\n

\\[\\left[ 126{,}133{,}140{,}147{,}154{,}161{,}168{,}175{,}182{,}189{,}196{,}203{,}210{,}217{,}224{,}231\\right] \\]<\/p>\n

\\[16\\]<\/p>\n

<\/p>\n

Este resultado lo podemos ver observando que hay floor(n\/m) n\u00famero divisibles por m menores que n<\/div>\n

<\/p>\n\n\n
(%i26) <\/span><\/td>\nfloor<\/span>(<\/span>235<\/span>\/<\/span>7<\/span>)<\/span>\u2212<\/span>floor<\/span>(<\/span>126<\/span>\/<\/span>7<\/span>)<\/span>;<\/span><\/span><\/td>\n<\/tr>\n<\/table>\n

\\[15\\]<\/p>\n

<\/p>\n

Observemos que no da el mismo resultado pues 126 es divisible por 7 y se resta dos veces. Para ser correcto ser\u00eda:<\/div>\n

<\/p>\n\n\n
(%i30) <\/span><\/td>\nini<\/span>:<\/span>126<\/span>$<\/span>
<\/span>fin<\/span>:<\/span>235<\/span>$<\/span>
<\/span>num<\/span>:<\/span>floor<\/span>(<\/span>fin<\/span>\/<\/span>7<\/span>)<\/span>\u2212<\/span>floor<\/span>(<\/span>ini<\/span>\/<\/span>7<\/span>)<\/span>$<\/span>
<\/span>if<\/span>(<\/span>mod<\/span>(<\/span>ini<\/span>,<\/span>7<\/span>)<\/span>=<\/span>0<\/span>)<\/span> then <\/span>print<\/span>(<\/span>num<\/span>+<\/span>1<\/span>)<\/span> else <\/span>print<\/span>(<\/span>num<\/span>)<\/span>$<\/span><\/span><\/td>\n<\/tr>\n<\/table>\n

16<\/p>\n

<\/p>\n

\n

1.5 Cifras de un n\u00famero en base 10<\/p>\n<\/div>\n

<\/p>\n

Veamos una curiosidad. Para conocer el n\u00famero de cifras que tiene un entero positivo cualquiera basta con calcular :\\[\\# n=\\text{N\u00ba cifras de }n=\\lfloor\\log(n)\\rfloor+1\\] donde \\(\\log(n)\\) es el logaritmo decimal.<\/div>\n

<\/p>\n

Ejercicio: \u00bfCu\u00e1ntas cifras tiene \\(F_{7}\\)?<\/div>\n

<\/p>\n\n\n
(%i32) <\/span><\/td>\nF7<\/span>:<\/span>2<\/span>^<\/span>(<\/span>2<\/span>^<\/span>7<\/span>)<\/span>+<\/span>1<\/span>;<\/span>
<\/span>print<\/span>(<\/span>\u00abn\u00famero de cifras: \u00ab<\/span>,<\/span>floor<\/span>(<\/span>log<\/span>(<\/span>F7<\/span>)<\/span>\/<\/span>log<\/span>(<\/span>10<\/span>)<\/span>)<\/span>+<\/span>1<\/span>)<\/span>$<\/span><\/span><\/td>\n<\/tr>\n<\/table>\n

n\u00famero de cifras: 39<\/p>\n

<\/p>\n

\n

2 Algoritmo de la numeracion<\/p>\n<\/div>\n

<\/p>\n

Aprenderemos a expresar un n\u00famero en base decimal a cualquier otra base<\/div>\n

<\/p>\n\n\n
(%i34) <\/span><\/td>\nF4<\/span>:<\/span>2<\/span>^<\/span>(<\/span>2<\/span>^<\/span>4<\/span>)<\/span>+<\/span>1<\/span>;<\/span>
<\/span>q<\/span>:<\/span>[<\/span>floor<\/span>(<\/span>F4<\/span>\/<\/span>7<\/span>)<\/span>,<\/span>mod<\/span>(<\/span>F4<\/span>,<\/span>7<\/span>)<\/span>]<\/span>;<\/span><\/span><\/td>\n<\/tr>\n<\/table>\n

\\[65537\\]<\/p>\n

\\[\\left[ 9362{,}3\\right] \\]<\/p>\n

<\/p>\n\n\n
(%i35) <\/span><\/td>\nq<\/span>:<\/span>[<\/span>floor<\/span>(<\/span>q<\/span>[<\/span>1<\/span>]<\/span>\/<\/span>7<\/span>)<\/span>,<\/span>mod<\/span>(<\/span>q<\/span>[<\/span>1<\/span>]<\/span>,<\/span>7<\/span>)<\/span>]<\/span>;<\/span><\/span><\/td>\n<\/tr>\n<\/table>\n

\\[\\left[ 1337{,}3\\right] \\]<\/p>\n

<\/p>\n\n\n
(%i36) <\/span><\/td>\nq<\/span>:<\/span>[<\/span>floor<\/span>(<\/span>q<\/span>[<\/span>1<\/span>]<\/span>\/<\/span>7<\/span>)<\/span>,<\/span>mod<\/span>(<\/span>q<\/span>[<\/span>1<\/span>]<\/span>,<\/span>7<\/span>)<\/span>]<\/span>;<\/span><\/span><\/td>\n<\/tr>\n<\/table>\n

\\[\\left[ 191{,}0\\right] \\]<\/p>\n

<\/p>\n\n\n
(%i37) <\/span><\/td>\nq<\/span>:<\/span>[<\/span>floor<\/span>(<\/span>q<\/span>[<\/span>1<\/span>]<\/span>\/<\/span>7<\/span>)<\/span>,<\/span>mod<\/span>(<\/span>q<\/span>[<\/span>1<\/span>]<\/span>,<\/span>7<\/span>)<\/span>]<\/span>;<\/span><\/span><\/td>\n<\/tr>\n<\/table>\n

\\[\\left[ 27{,}2\\right] \\]<\/p>\n

<\/p>\n\n\n
(%i38) <\/span><\/td>\nq<\/span>:<\/span>[<\/span>floor<\/span>(<\/span>q<\/span>[<\/span>1<\/span>]<\/span>\/<\/span>7<\/span>)<\/span>,<\/span>mod<\/span>(<\/span>q<\/span>[<\/span>1<\/span>]<\/span>,<\/span>7<\/span>)<\/span>]<\/span>;<\/span><\/span><\/td>\n<\/tr>\n<\/table>\n

\\[\\left[ 3{,}6\\right] \\]<\/p>\n

<\/p>\n\n\n
(%i39) <\/span><\/td>\nq<\/span>:<\/span>[<\/span>floor<\/span>(<\/span>q<\/span>[<\/span>1<\/span>]<\/span>\/<\/span>7<\/span>)<\/span>,<\/span>mod<\/span>(<\/span>q<\/span>[<\/span>1<\/span>]<\/span>,<\/span>7<\/span>)<\/span>]<\/span>;<\/span><\/span><\/td>\n<\/tr>\n<\/table>\n

\\[\\left[ 0{,}3\\right] \\]<\/p>\n

<\/p>\n

Terminamos porque hemos llegado a un cociente igual a 0, el numero que buscamos es (362033) en base 7<\/div>\n

<\/p>\n

\n

2.1 Algoritmo<\/h3>\n<\/div>\n

<\/p>\n

El proceso unificado es:<\/div>\n

<\/p>\n\n\n
(%i44) <\/span><\/td>\nF4<\/span>:<\/span>2<\/span>^<\/span>(<\/span>2<\/span>^<\/span>4<\/span>)<\/span>+<\/span>1<\/span>$<\/span>
<\/span>q<\/span>:<\/span>[<\/span>floor<\/span>(<\/span>F4<\/span>\/<\/span>7<\/span>)<\/span>,<\/span>mod<\/span>(<\/span>F4<\/span>,<\/span>7<\/span>)<\/span>]<\/span>$<\/span>
<\/span>m<\/span>:<\/span>matrix<\/span>(<\/span>[<\/span>q<\/span>[<\/span>1<\/span>]<\/span>]<\/span>,<\/span>[<\/span>q<\/span>[<\/span>2<\/span>]<\/span>]<\/span>)<\/span>$<\/span>
<\/span>for <\/span>i<\/span>:<\/span>1<\/span> thru <\/span>5<\/span> do <\/span>(<\/span>
<\/span> \u00a0\u00a0 q<\/span>:<\/span>[<\/span>floor<\/span>(<\/span>q<\/span>[<\/span>1<\/span>]<\/span>\/<\/span>7<\/span>)<\/span>,<\/span>mod<\/span>(<\/span>q<\/span>[<\/span>1<\/span>]<\/span>,<\/span>7<\/span>)<\/span>]<\/span>,<\/span>
<\/span> \u00a0\u00a0 m<\/span>:<\/span>addcol<\/span>(<\/span>m<\/span>,<\/span>transpose<\/span>(<\/span>q<\/span>)<\/span>)<\/span>
<\/span>)<\/span>$<\/span>
<\/span>m<\/span>;<\/span><\/span><\/td>\n<\/tr>\n<\/table>\n

\\[\\begin{pmatrix}9362 & 1337 & 191 & 27 & 3 & 0\\\\3 & 3 & 0 & 2 & 6 & 3\\end{pmatrix}\\]<\/p>\n

<\/p>\n

Recordemos que vimos que los d\u00edgitos de un n\u00famero se calculaban con floor(log(n)\/log(10))+1, pues los d\u00edgitos en cualquier base b es floor(log(n)\/log(b))+1, por tanto en el for anterior basta con hacer el thru hasta floor(log(F4)\/log(7))<\/div>\n

<\/p>\n

\n

2.1.1 Ejercicio<\/h4>\n<\/div>\n

<\/p>\n

\u00bfCu\u00e1nto suman los d\u00edgitos de F4 en base 7?<\/div>\n

<\/p>\n\n\n
(%i45) <\/span><\/td>\nsum<\/span>(<\/span>m<\/span>[<\/span>2<\/span>,<\/span>i<\/span>]<\/span>,<\/span>i<\/span>,<\/span>1<\/span>,<\/span>6<\/span>)<\/span>;<\/span><\/span><\/td>\n<\/tr>\n<\/table>\n

\\[17\\]<\/p>\n

<\/p>\n

\n

2.1.2 Ejercicio<\/h4>\n<\/div>\n

<\/p>\n

Expresar F5 en base 8<\/div>\n

<\/p>\n\n\n
(%i50) <\/span><\/td>\nF5<\/span>:<\/span>2<\/span>^<\/span>(<\/span>2<\/span>^<\/span>5<\/span>)<\/span>+<\/span>1<\/span>$<\/span>
<\/span>q<\/span>:<\/span>floor<\/span>(<\/span>F5<\/span>\/<\/span>8<\/span>)<\/span>$<\/span>
<\/span>d<\/span>:<\/span>[<\/span>mod<\/span>(<\/span>F5<\/span>,<\/span>8<\/span>)<\/span>]<\/span>$<\/span>
<\/span>while<\/span>(<\/span>q<\/span>><\/span>0<\/span>)<\/span> do<\/span>(<\/span>
<\/span> \u00a0\u00a0 d<\/span>:<\/span>append<\/span>(<\/span>d<\/span>,<\/span>[<\/span>mod<\/span>(<\/span>q<\/span>,<\/span>8<\/span>)<\/span>]<\/span>)<\/span>,<\/span>
<\/span> \u00a0\u00a0 q<\/span>:<\/span>floor<\/span>(<\/span>q<\/span>\/<\/span>8<\/span>)<\/span>
<\/span> \u00a0\u00a0 )<\/span>$<\/span>
<\/span>concat<\/span>(<\/span>F5<\/span>,<\/span> es (<\/span>,<\/span>sum<\/span>(<\/span>d<\/span>[<\/span>i<\/span>]<\/span>\u00b7<\/span>10<\/span>^<\/span>(<\/span>i<\/span>\u2212<\/span>1<\/span>)<\/span>,<\/span>i<\/span>,<\/span>1<\/span>,<\/span>length<\/span>(<\/span>d<\/span>)<\/span>)<\/span>,<\/span>\u00ab)_\u00bb<\/span>,<\/span>8<\/span>)<\/span>;<\/span><\/span><\/td>\n<\/tr>\n<\/table>\n

\\[4294967297 \\mbox{ es } (40000000001)_8\\]<\/p>\n

<\/p>\n

\n

2.2 Funci\u00f3n algoritmo<\/h3>\n<\/div>\n

<\/p>\n

Veamos una funci\u00f3n que nos de la expresi\u00f3n de cualquier n\u00famero en base 10 a base b.<\/div>\n

<\/p>\n\n\n
(%i52) <\/span><\/td>\nbase<\/span>(<\/span>n<\/span>,<\/span>b<\/span>)<\/span>:<\/span>=<\/span>block<\/span>(<\/span>[<\/span>q<\/span>,<\/span>d<\/span>]<\/span>,<\/span>
<\/span> \u00a0\u00a0 q<\/span>:<\/span>floor<\/span>(<\/span>n<\/span>\/<\/span>b<\/span>)<\/span>,<\/span>
<\/span> \u00a0\u00a0 d<\/span>:<\/span>[<\/span>mod<\/span>(<\/span>n<\/span>,<\/span>b<\/span>)<\/span>]<\/span>,<\/span>
<\/span> \u00a0\u00a0 while<\/span>(<\/span>q<\/span>><\/span>0<\/span>)<\/span> do <\/span>(<\/span>
<\/span> \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 d<\/span>:<\/span>append<\/span>(<\/span>d<\/span>,<\/span>[<\/span>mod<\/span>(<\/span>q<\/span>,<\/span>b<\/span>)<\/span>]<\/span>)<\/span>,<\/span>
<\/span> \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 q<\/span>:<\/span>floor<\/span>(<\/span>q<\/span>\/<\/span>b<\/span>)<\/span>
<\/span> \u00a0\u00a0 )<\/span>,<\/span>
<\/span> \u00a0\u00a0 d<\/span>
<\/span>)<\/span>$<\/span>\/* Esta funci\u00f3n nos da los d\u00edgitos*\/<\/span>
<\/span>num_base<\/span>(<\/span>n<\/span>,<\/span>b<\/span>)<\/span>:<\/span>=<\/span>block<\/span>(<\/span>[<\/span>d<\/span>,<\/span>i<\/span>,<\/span>p<\/span>]<\/span>,<\/span>
<\/span> \u00a0\u00a0 d<\/span>:<\/span>base<\/span>(<\/span>n<\/span>,<\/span>b<\/span>)<\/span>,<\/span>
<\/span> \u00a0\u00a0 p<\/span>:<\/span>sum<\/span>(<\/span>d<\/span>[<\/span>i<\/span>]<\/span>\u00b7<\/span>10<\/span>^<\/span>(<\/span>i<\/span>\u2212<\/span>1<\/span>)<\/span>,<\/span>i<\/span>,<\/span>1<\/span>,<\/span>length<\/span>(<\/span>d<\/span>)<\/span>)<\/span>,<\/span>
<\/span> \u00a0\u00a0 concat<\/span>(<\/span>\u00ab(\u00ab<\/span>,<\/span>p<\/span>,<\/span>\u00ab)_\u00bb<\/span>,<\/span>b<\/span>)<\/span>
<\/span>)<\/span>$<\/span>\/*Con esta funci\u00f3n lo expresamos en base b*\/<\/span><\/span><\/td>\n<\/tr>\n<\/table>\n

<\/p>\n\n\n
(%i53) <\/span><\/td>\nprint<\/span>(<\/span>F4<\/span>,<\/span>\u00ab=\u00bb<\/span>,<\/span>num_base<\/span>(<\/span>F4<\/span>,<\/span>7<\/span>)<\/span>)<\/span>$<\/span><\/span><\/td>\n<\/tr>\n<\/table>\n

\\[65537=(362033)_7\\]<\/p>\n

<\/p>\n\n\n
(%i55) <\/span><\/td>\nF6<\/span>:<\/span>2<\/span>^<\/span>(<\/span>2<\/span>^<\/span>6<\/span>)<\/span>+<\/span>1<\/span>$<\/span>print<\/span>(<\/span>num_base<\/span>(<\/span>F6<\/span>,<\/span>8<\/span>)<\/span>)<\/span>$<\/span><\/span><\/td>\n<\/tr>\n<\/table>\n

\\[(2000000000000000000001)_8\\]<\/p>\n

<\/p>\n\n\n
(%i57) <\/span><\/td>\nF7<\/span>:<\/span>2<\/span>^<\/span>(<\/span>2<\/span>^<\/span>7<\/span>)<\/span>+<\/span>1<\/span>$<\/span>print<\/span>(<\/span>num_base<\/span>(<\/span>F7<\/span>,<\/span>8<\/span>)<\/span>)<\/span>$<\/span><\/span><\/td>\n<\/tr>\n<\/table>\n

\\[(4000000000000000000000000000000000000000001)_8\\]<\/p>\n","protected":false},"excerpt":{"rendered":"

1. El algoritmo de la divisi\u00f3n Dados dos n\u00fameros enteros \\(a\\) y \\(b\\), con \\(a\\) no nulo, la divisi\u00f3n eucl\u00eddea asocia un cociente \\(q\\in\\mathbb{Z}\\) y un resto \\(r\\in\\mathbb{Z}\\), \u00fanicos, que verifican: \\[b=q\\,a+r,\\quad…<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[8],"tags":[9],"class_list":["post-759","post","type-post","status-publish","format-standard","hentry","category-matematica-discreta","tag-practicas-mad"],"_links":{"self":[{"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=\/wp\/v2\/posts\/759","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=759"}],"version-history":[{"count":8,"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=\/wp\/v2\/posts\/759\/revisions"}],"predecessor-version":[{"id":771,"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=\/wp\/v2\/posts\/759\/revisions\/771"}],"wp:attachment":[{"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=759"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=759"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=759"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}