chore(i8n,learn): processed translations

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

@@ -1,48 +1,48 @@
 ---
 id: 5900f4ab1000cf542c50ffbd
-title: 问题318：2011个九
+title: 'Problem 318: 2011 nines'
 challengeType: 5
-videoUrl: ''
+forumTopicId: 301974
 dashedName: problem-318-2011-nines
 ---
 
 # --description--
 
-考虑实数√2+√3。
+Consider the real number √2+√3.
 
-当我们计算√2+√3的偶数幂时
+When we calculate the even powers of √2+√3
 
-我们得到：
+we get:
 
-（√2+√3）2 = 9.898979485566356 ...
+(√2+√3)2 = 9.898979485566356...
 
-（√2+√3）4 = 97.98979485566356 ...
+(√2+√3)4 = 97.98979485566356...
 
-（√2+√3）6 = 969.998969071069263 ...
+(√2+√3)6 = 969.998969071069263...
 
-（√2+√3）8 = 9601.99989585502907 ...
+(√2+√3)8 = 9601.99989585502907...
 
-（√2+√3）10 = 95049.999989479221 ...
+(√2+√3)10 = 95049.999989479221...
 
-（√2+√3）12 = 940897.9999989371855 ...
+(√2+√3)12 = 940897.9999989371855...
 
-（√2+√3）14 = 9313929.99999989263 ...
+(√2+√3)14 = 9313929.99999989263...
 
-（√2+√3）16 = 92198401.99999998915 ...
+(√2+√3)16 = 92198401.99999998915...
 
-这些幂的小数部分开头的连续九个数字似乎没有减少。 实际上，可以证明（√2+√3）2n的小数部分对于大n接近1。
+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.
 
-考虑形式为√p+√q的所有实数，其中p和q为正整数，且p <q，使得小数部分 （√p+√q）的2n对于大n接近1。
+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.
 
-令C（p，q，n）为（√p+√q）2n的小数部分开头的连续九个数字。
+Let C(p,q,n) be the number of consecutive nines at the beginning of the fractional part of (√p+√q)2n.
 
-令N（p，q）为n的最小值，以使C（p，q，n）≥2011。
+Let N(p,q) be the minimal value of n such that C(p,q,n) ≥ 2011.
 
-求p + q≤2011的∑N（p，q）。
+Find ∑N(p,q) for p+q ≤ 2011.
 
 # --hints--
 
-`euler318()`应该返回709313889。
+`euler318()` should return 709313889.
 
 ```js
 assert.strictEqual(euler318(), 709313889);