chore(i18n,curriculum): processed translations (#44221)

curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-318-2011-nines.md

@@ -1,6 +1,6 @@
 ---
 id: 5900f4ab1000cf542c50ffbd
-title: 'Problem 318: 2011 nines'
+title: 'Problema 318: 2011 noves'
 challengeType: 5
 forumTopicId: 301974
 dashedName: problem-318-2011-nines
@@ -8,44 +8,28 @@ dashedName: problem-318-2011-nines
 
 # --description--
 
-Consider the real number √2+√3.
+Considere o número real $\sqrt{2} + \sqrt{3}$.
 
-When we calculate the even powers of √2+√3
+Quando calculamos as potências pares de $\sqrt{2} + \sqrt{3}$ obtemos:
 
-we get:
+$$\begin{align} & {(\sqrt{2} + \sqrt{3})}^2 = 9.898979485566356\ldots \\\\ & {(\sqrt{2} + \sqrt{3})}^4 = 97.98979485566356\ldots \\\\ & {(\sqrt{2} + \sqrt{3})}^6 = 969.998969071069263\ldots \\\\ & {(\sqrt{2} + \sqrt{3})}^8 = 9601.99989585502907\ldots \\\\ & {(\sqrt{2} + \sqrt{3})}^{10} = 95049.999989479221\ldots \\\\ & {(\sqrt{2} + \sqrt{3})}^{12} = 940897.9999989371855\ldots \\\\ & {(\sqrt{2} + \sqrt{3})}^{14} = 9313929.99999989263\ldots \\\\ & {(\sqrt{2} + \sqrt{3})}^{16} = 92198401.99999998915\ldots \\\\ \end{align}$$
 
-(√2+√3)2 = 9.898979485566356...
+Parece que o número de noves consecutivos no início da parte fracionária dessas potências não diminui. Na verdade, pode ser provado que a parte fracionária de ${(\sqrt{2} + \sqrt{3})}^{2n}$ aproxima-se de 1 para $n$ grandes.
 
-(√2+√3)4 = 97.98979485566356...
+Considere todos os números reais da forma $\sqrt{p} + \sqrt{q}$ com $p$ e $q$ números inteiros positivos e $p < q$, tal que a parte fracionária de ${(\sqrt{p} + \sqrt{q})}^{2n}$ se aproxima de 1 para $n$ grandes.
 
-(√2+√3)6 = 969.998969071069263...
+Considere $C(p,q,n)$ como o número de noves consecutivos no início da parte fracionária de ${(\sqrt{p} + \sqrt{q})}^{2n}$.
 
-(√2+√3)8 = 9601.99989585502907...
+Considere $N(p,q)$ como o valor mínimo de $n$, tal que $C(p,q,n) ≥ 2011$.
 
-(√2+√3)10 = 95049.999989479221...
-
-(√2+√3)12 = 940897.9999989371855...
-
-(√2+√3)14 = 9313929.99999989263...
-
-(√2+√3)16 = 92198401.99999998915...
-
-It looks like that the number of consecutive nines at the beginning of the fractional part of these powers is non-decreasing. In fact it can be proven that the fractional part of (√2+√3)2n approaches 1 for large n.
-
-Consider all real numbers of the form √p+√q with p and q positive integers and p<q, such that the fractional part of (√p+√q)2n approaches 1 for large n.
-
-Let C(p,q,n) be the number of consecutive nines at the beginning of the fractional part of (√p+√q)2n.
-
-Let N(p,q) be the minimal value of n such that C(p,q,n) ≥ 2011.
-
-Find ∑N(p,q) for p+q ≤ 2011.
+Encontre $\sum N(p,q)$ para $p + q ≤ 2011$.
 
 # --hints--
 
-`euler318()` should return 709313889.
+`twoThousandElevenNines()` deve retornar `709313889`.
 
 ```js
-assert.strictEqual(euler318(), 709313889);
+assert.strictEqual(twoThousandElevenNines(), 709313889);
 ```
 
 # --seed--
@@ -53,12 +37,12 @@ assert.strictEqual(euler318(), 709313889);
 ## --seed-contents--
 
 ```js
-function euler318() {
+function twoThousandElevenNines() {
 
   return true;
 }
 
-euler318();
+twoThousandElevenNines();
 ```
 
 # --solutions--