import{E as A,c as o,aO as P,h as y,r as $,cC as i,w as a,b5 as l,cD as f,a0 as w,m as C,l as Q}from"./index-XCg2QAX4.js";import{l as S,k as b,w as G}from"./message-nJyRkpoH.js";import{c as I}from"./modales-zNa05bdf.js";import{s as M}from"./macroSvgJs-z7xlNAOz.js";const q="Solve a gears exercise",U="01/04/2023",B="ce352",z="3A12";function E(){A.call(this),this.titre=q,this.consigne="",o.isHtml?this.spacing=2:this.spacing=2,o.isHtml?this.spacingCorr=2:this.spacingCorr=1,this.nbQuestionsModifiable=!1,this.nbQuestions=4,this.nbCols=1,this.nbColsCorr=1,this.listePackages="bclogo",this.sup=!1;const k="3A12";this.nouvelleVersion=function(h){let x;o.isHtml&&(this.boutonAide=I(h,"assets/pdf/FicheArithmetique-3A13.pdf","Cheat sheet - Arithmetic (Sébastien Lozano)","Cheat sheet")),this.listeQuestions=[],this.listeCorrections=[],this.contenu="",this.contenuCorrection="";const T=P([1,2,3],this.nbQuestions);let g="Gearbox, bicycle transmission, motorcycle transmission, electric drill, all of this works with gears! But actually, how do gears work?";if(o.isHtml){const r=`divIntro${`${k}_${Date.now()}`}`;M(),g+=G(`Please note, the wheels below do not have the number of teeth shown! <br> <svg-gear id="${r}"></svg-gear>`,"numbers","Help")}else g+="\\\\ \\textit{Attention, the wheels below do not have the number of teeth in the statement!} \\\\ \\Gears[Color=white,Unit=1mm]{1/24, 1/9}";this.introduction=S({titre:"Gear arithmetic",texte:g,couleur:"numbers"});for(let u=0,r,s,c,D=0;u<this.nbQuestions&&D<50;){x=T[u];let e,t,m=`Given two integers a and b, when the smallest common multiple of $a$ and $b$ is worth $a \\times b$ ${o.isHtml?"":"\\\\"}( $ppcm(a,b)=a\\times b$ ), we say that`;o.isHtml?m+="<b>the numbers a and b are relatively prime.</b>":m+="$\\textbf{the numbers a and b are relatively prime}$.";let d=`Given two integers a and b, when the largest divisor common to $a$ and $b$ is worth $1$ ${o.isHtml?"":"\\\\"} ( $gcd(a,b)=1$ ), we say that`;o.isHtml?d+="<b>the numbers a and b are relatively prime.</b>":d+="$\\textbf{the numbers a and b are relatively prime}$.";let p="Given two integers a and b, when $a$ and $b$ have no other common divisor than $1$, we say that";switch(o.isHtml?p+="<b>the numbers a and b are relatively prime.</b>":p+="$\\textbf{the numbers a and b are relatively prime}$.",x){case 1:{e=$(5,30),t=$(5,30,e),r=`Wheel n$\\degree$1 has $${e}$ teeth and wheel n$\\degree$2 has $${t}$ teeth.`,r+="<br>"+l(0)+` Write the list of multiples of $${e}$ and $${t}$ until you find a common multiple.`,i(e,t)===e*t&&(r+=`<br>Justify that ${e} and ${t} are`,r+=b(h+1,1,`prime numbers among themselves${o.isHtml?"?":""}`,"Definition from the least common multiple",`${o.isHtml?"<br>":"\\\\"} ${m}`)),r+=o.isHtml?"":"?",r+="<br>"+l(1)+" Deduce the number of revolutions of each wheel before returning to their initial position.",s=l(0)+` List of first multiples of $${e}$: <br>`;let v=5-i(e,t)/e%5,H=i(e,t)/e+v;for(let n=1;n<H+1;n++)s+=`$${n}\\times${e} =`,n===i(e,t)/e?(s+=C(a(n*e)),s+="$ ;"):s+=`${a(n*e)}$ ;`,n%5===0&&(s+="<br>");s+="$\\ldots$",s+="<br>",s+=` List of first multiples of $${t}$: <br>`,v=5-i(e,t)/t%5,H=i(e,t)/t+v;for(let n=1;n<H+1;n++)s+=`$${n}\\times${t} =`,n===i(e,t)/t?(s+=C(a(n*t)),s+="$ ;"):s+=`${a(n*t)}$ ;`,n%5===0&&(s+="<br>");s+="$\\ldots$",s+="<br>",i(e,t)===e*t&&(s+="$ppcm("+e+";"+t+")="+e+"\\times"+t+`$ therefore $${e}$ and $${t}$ are`,s+=b(h+2,1,`prime numbers among themselves${o.isHtml?".":""}`,"Definition from the least common multiple",`${o.isHtml?"<br>":"\\\\"} ${m}`)),s+=o.isHtml?"":".",s+="<br>",s+=l(1)+` The smallest multiple common to $${e}$ and $${t}$ is therefore worth $${i(e,t)}$.<br>It is therefore sufficient for each wheel to turn $${i(e,t)}$ teeth to make a whole number of revolutions and thus return to its initial position.<br>In fact, each wheel must rotate so that the total number of teeth used is a multiple of its number of teeth or at least $${a(i(e,t))}$ teeth.`,s+=`<br> This corresponds to $(${i(e,t)}\\text{ teeth})\\div (${e}\\text{ teeth/revolution}) = ${i(e,t)/e}$`,i(e,t)/e===1?s+=" round":s+=" towers ",s+="for wheel n$\\degree$1.",s+=`<br>This corresponds to $(${i(e,t)}\\text{ teeth})\\div (${t}\\text{ teeth/revolution}) = ${i(e,t)/t}$`,i(e,t)/t===1?s+=" round":s+=" towers ",s+="for wheel n$\\degree$2."}break;case 2:if(this.sup)for(e=$(51,100),t=$(51,100,e);t%e===0||e%t===0;)t=$(51,100,e);else for(e=$(31,80),t=$(31,80,e);t%e===0||e%t===0;)t=$(51,100,e);r=`Wheel n$\\degree$1 has $${e}$ teeth and wheel n$\\degree$2 has $${t}$ teeth.`,r+="<br>"+l(0)+` Decompose $${e}$ and $${t}$ into product of prime factors.`,i(e,t)===e*t&&(r+=`<br>Justify that ${e} and ${t} are`,r+=b(h+3,1,`prime numbers among themselves${o.isHtml?"?":""}`,"Three equivalent definitions to choose from",`<br>- ${m} ${o.isHtml?"<br>-":"\\\\-"} ${d} ${o.isHtml?"<br>-":"\\\\-"} ${p}`)),r+=o.isHtml?"":"?",r+="<br>"+l(1)+" Deduce the number of revolutions of each wheel before returning to their initial position.",s="For a higher number of teeth, it is more convenient to use product decompositions of prime factors.",s+="<br>"+l(0)+` Decomposition of $${e}$ into product of prime factors: $${e} = ${f(e)}$.`,s+=`<br> Decomposition of $${t}$ into product of prime factors: $${t} = ${f(t)}$.`,s+="<br>",i(e,t)===e*t&&(s+="Proposal of three valid corrections for the deduction: <br>",s+="Correction proposal 1: <br>",s+=`According to the previous calculations, $ppcm(${e},${t})= ${f(i(e,t))}$.<br>`,s+=`So $${e}$ and $${t}$ are`,s+=b(h+4,1,`prime numbers among themselves${o.isHtml?".":""}`,"Definition from the least common multiple",`${o.isHtml?"<br>":"\\\\"} ${m}`)),s+=o.isHtml?"":".",w(e,t)===1&&(s+="<br>Correction proposal 2: <br>",s+=`According to the previous calculations, $gcd(${e},${t})= ${w(e,t)===1?1:""} ${f(w(e,t))}$.<br>`,s+=`So $${e}$ and $${t}$ are`,s+=b(h+5,1,`prime numbers among themselves${o.isHtml?".":""}`,"Definition from the greatest common divisor",`${o.isHtml?"<br>":"\\\\"} ${d}`)),s+=o.isHtml?"":".",w(e,t)===1&&(s+="<br>Correction proposal 3: <br>",s+=`According to the previous calculations, the only divisor common to $${e}$ and $${t}$ is worth $1$.<br>`,s+=`So $${e}$ and $${t}$ are`,s+=b(h+6,1,`prime numbers among themselves${o.isHtml?".":""}`,"Definition from the intersection of common divisors",`${o.isHtml?"<br>":"\\\\"} ${p}`)),s+=o.isHtml?"":".",s+="<br>",s+=l(1)+` To find the initial position, each wheel must rotate so that the total number of teeth used is a multiple of its number of teeth.<br>Or, thanks to the previous decompositions, at least $${f(i(e,t))} = ${i(e,t)}$ teeth.`,s+=`<br> This corresponds to $(${a(i(e,t))}\\text{ teeth})\\div (${e}\\text{ teeth/revolution}) = ${i(e,t)/e}$`,i(e,t)/e===1?s+=" round":s+=" towers ",s+="for wheel n$\\degree$1.",s+=`<br> This corresponds to $(${a(i(e,t))}\\text{ teeth})\\div (${t}\\text{ teeth/revolution}) = ${i(e,t)/t}$`,i(e,t)/t===1?s+=" round":s+=" towers ",s+="for wheel n$\\degree$2.";break;case 3:this.sup?c=y([2,3,4,5,6]):c=y([2,3]),e=$(5,15),t=$(5,80,e),e*=c,t*=c,r=`Wheel n$\\degree$2 now has $${t}$ teeth.`,r+=` Determine the number of teeth of the wheel n$\\degree$1 which would make $${i(e,t)/e}$`,i(e,t)/e===1?r+=" round":r+=" towers ",r+=` while the wheel n$\\degree$2 makes it $${i(e,t)/t}$.`,s=`Since the wheel n$\\degree$2, which has $${t}$ teeth, makes $${i(e,t)/t}$`,i(e,t)/t===1?s+=" round":s+=" towers ",s+=`, this represents $${a(i(e,t))}$ teeth.`,s+=`<br>The wheel n$\\degree$1 must therefore also rotate by $${a(i(e,t))}$ teeth, this in $${i(e,t)/e}$`,i(e,t)/e===1?s+=" round":s+=" towers ",s+=".",s+=`<br> We therefore obtain $(${a(i(e,t))}\\text{ teeth})\\div (${i(e,t)/e}\\text{`,i(e,t)/e===1?s+=" round":s+=" towers ",s+=`}) = ${e} \\text{ teeth/revolution}.$`,s+=`<br>The wheel n$\\degree$1 therefore has $${e}$ teeth.`;break}this.questionJamaisPosee(u,e,t)&&(this.listeQuestions.push(r),this.listeCorrections.push(s),u++),D++}Q(this)}}export{U as dateDeModifImportante,E 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