chore(i8n,learn): processed translations

curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-21-amicable-numbers.md

@@ -1,36 +1,48 @@
 ---
 id: 5900f3811000cf542c50fe94
-title: 问题21：友好的数字
+title: 'Problem 21: Amicable numbers'
 challengeType: 5
-videoUrl: ''
+forumTopicId: 301851
 dashedName: problem-21-amicable-numbers
 ---
 
 # --description--
 
-设d（ `n` ）定义为`n`的适当除数之`和` （小于`n的`数均匀分成`n` ）。如果d（ `a` ）= `b`并且d（ `b` ）= `a` ，其中`a` ≠ `b` ，则`a`和`b`是友好对，并且`a`和`b`中的每`一个`被称为友好数字。例如，220的适当除数是1,2,4,5,10,11,20,22,44,55和110;因此d（220）= 284. 284的适当除数是1,2,4,71和142;所以d（284）= 220.评估`n`下所有友好数字的总和。
+Let d(`n`) be defined as the sum of proper divisors of `n` (numbers less than `n` which divide evenly into `n`).
+
+If d(`a`) = `b` and d(`b`) = `a`, where `a` ≠ `b`, then `a` and `b` are an amicable pair and each of `a` and `b` are called amicable numbers.
+
+For example, the proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20, 22, 44, 55 and 110; therefore d(220) = 284. The proper divisors of 284 are 1, 2, 4, 71 and 142; so d(284) = 220.
+
+Evaluate the sum of all the amicable numbers under `n`.
 
 # --hints--
 
-`sumAmicableNum(1000)`应返回504。
+`sumAmicableNum(1000)` should return a number.
+
+```js
+assert(typeof sumAmicableNum(1000) === 'number');
+```
+
+`sumAmicableNum(1000)` should return 504.
 
 ```js
 assert.strictEqual(sumAmicableNum(1000), 504);
 ```
 
-`sumAmicableNum(2000)`应该返回2898。
+`sumAmicableNum(2000)` should return 2898.
 
 ```js
 assert.strictEqual(sumAmicableNum(2000), 2898);
 ```
 
-`sumAmicableNum(5000)`应该返回8442。
+`sumAmicableNum(5000)` should return 8442.
 
 ```js
 assert.strictEqual(sumAmicableNum(5000), 8442);
 ```
 
-`sumAmicableNum(10000)`应返回31626。
+`sumAmicableNum(10000)` should return 31626.
 
 ```js
 assert.strictEqual(sumAmicableNum(10000), 31626);