chore(i8n,learn): processed translations

curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-30-digit-n-powers.md

@@ -1,48 +1,54 @@
 ---
 id: 5900f38a1000cf542c50fe9d
-title: 问题30：数字n次方
+title: 'Problem 30: Digit n powers'
 challengeType: 5
-videoUrl: ''
+forumTopicId: 301953
 dashedName: problem-30-digit-n-powers
 ---
 
 # --description--
 
-令人惊讶的是，只有三个数字可以写为它们的数字的四次幂之和：
+Surprisingly there are only three numbers that can be written as the sum of fourth powers of their digits:
 
-1634 = 1 <sup>4</sup> + 6 <sup>4</sup> + 3 <sup>4</sup> + 4 <sup>4</sup>
+<div style='margin-left: 4em;'>
+  1634 = 1<sup>4</sup> + 6<sup>4</sup> + 3<sup>4</sup> + 4<sup>4</sup><br>
+  8208 = 8<sup>4</sup> + 2<sup>4</sup> + 0<sup>4</sup> + 8<sup>4</sup><br>
+  9474 = 9<sup>4</sup> + 4<sup>4</sup> + 7<sup>4</sup> + 4<sup>4</sup><br>
+</div>
 
-8208 = 8 <sup>4</sup> + 2 <sup>4</sup> + 0 <sup>4</sup> + 8 <sup>4</sup>
+As 1 = 1<sup>4</sup> is not a sum it is not included.
 
-9474 = 9 <sup>4</sup> + 4 <sup>4</sup> + 7 <sup>4</sup> + 4 <sup>4</sup>
+The sum of these numbers is 1634 + 8208 + 9474 = 19316.
 
-由于1 = 1 <sup>4</sup>不是总和，因此不包括在内。
-
-这些数字的总和为1634 + 8208 + 9474 = 19316。
-
-找出所有可以写为数字n次幂的数字之和。
+Find the sum of all the numbers that can be written as the sum of `n` powers of their digits.
 
 # --hints--
 
-`digitnPowers(2)`应该返回0。
+`digitnPowers(2)` should return a number.
+
+```js
+assert(typeof digitnPowers(2) === 'number');
+```
+
+`digitnPowers(2)` should return 0.
 
 ```js
 assert(digitnPowers(2) == 0);
 ```
 
-`digitnPowers(3)`应该返回1301。
+`digitnPowers(3)` should return 1301.
 
 ```js
 assert(digitnPowers(3) == 1301);
 ```
 
-`digitnPowers(4)`应该返回19316。
+`digitnPowers(4)` should return 19316.
 
 ```js
 assert(digitnPowers(4) == 19316);
 ```
 
-`digitnPowers(5)`应该返回443839。
+`digitnPowers(5)` should return 443839.
 
 ```js
 assert(digitnPowers(5) == 443839);