chore(i8n,learn): processed translations

curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-27-quadratic-primes.md

@@ -1,40 +1,58 @@
 ---
 id: 5900f3871000cf542c50fe9a
-title: 问题27：二次素数
+title: 'Problem 27: Quadratic primes'
 challengeType: 5
-videoUrl: ''
+forumTopicId: 301919
 dashedName: problem-27-quadratic-primes
 ---
 
 # --description--
 
-欧拉发现了显着的二次公式：$ n ^ 2 + n + 41 $事实证明，公式将为连续的整数值$ 0 \\ le n \\ le 39 $产生40个素数。但是，当$ n = 40时，40 ^ 2 + 40 + 41 = 40（40 + 1）+ 41 $可被41整除，当然$ n = 41时，41 ^ 2 + 41 + 41 $显然可以被整除41.发现了令人难以置信的公式$ n ^ 2 - 79n + 1601 $，它为连续值$ 0 \\ le n \\ le 79 $产生80个素数。系数-79和1601的乘积是-126479。考虑形式的二次方：
+Euler discovered the remarkable quadratic formula:
 
-$ n ^ 2 + an + b $，其中$ | a | <range $和$ | b | \\ le $ $其中$ | n | $是$ n $的模数/绝对值，例如$ | 11 | = 11 $和$ | -4 | = 4 $
+<div style='margin-left: 4em;'>$n^2 + n + 41$</div>
 
-找到系数的乘积，$ a $和$ b $，用于生成连续值$ n $的最大素数数的二次表达式，从$ n = 0 $开始。
+It turns out that the formula will produce 40 primes for the consecutive integer values $0 \\le n \\le 39$. However, when $n = 40, 40^2 + 40 + 41 = 40(40 + 1) + 41$ is divisible by 41, and certainly when $n = 41, 41^2 + 41 + 41$ is clearly divisible by 41.
+
+The incredible formula $n^2 - 79n + 1601$ was discovered, which produces 80 primes for the consecutive values $0 \\le n \\le 79$. The product of the coefficients, −79 and 1601, is −126479.
+
+Considering quadratics of the form:
+
+<div style='margin-left: 4em;'>
+  $n^2 + an + b$, where $|a| < range$ and $|b| \le range$<br>
+  where $|n|$ is the modulus/absolute value of $n$<br>
+  e.g. $|11| = 11$ and $|-4| = 4$<br>
+</div>
+
+Find the product of the coefficients, $a$ and $b$, for the quadratic expression that produces the maximum number of primes for consecutive values of $n$, starting with $n = 0$.
 
 # --hints--
 
-`quadraticPrimes(200)`应返回-4925。
+`quadraticPrimes(200)` should return a number.
+
+```js
+assert(typeof quadraticPrimes(200) === 'number');
+```
+
+`quadraticPrimes(200)` should return -4925.
 
 ```js
 assert(quadraticPrimes(200) == -4925);
 ```
 
-`quadraticPrimes(500)`应返回-18901。
+`quadraticPrimes(500)` should return -18901.
 
 ```js
 assert(quadraticPrimes(500) == -18901);
 ```
 
-`quadraticPrimes(800)`应返回-43835。
+`quadraticPrimes(800)` should return -43835.
 
 ```js
 assert(quadraticPrimes(800) == -43835);
 ```
 
-`quadraticPrimes(1000)`应返回-59231。
+`quadraticPrimes(1000)` should return -59231.
 
 ```js
 assert(quadraticPrimes(1000) == -59231);