From c0aaecba634d707a49244ae27290f488c405dddc Mon Sep 17 00:00:00 2001
From: gikf <60067306+gikf@users.noreply.github.com>
Date: Tue, 1 Jun 2021 08:50:51 +0200
Subject: [PATCH] fix(curriculum): rework Project Euler 72 (#42282)

* fix: rework challenge to use argument in function

* fix: add solution

* fix: use MathJax in fractions and equations

* fix: missing backticks
---
 .../problem-72-counting-fractions.md          | 60 +++++++++++++++----
 1 file changed, 49 insertions(+), 11 deletions(-)

diff --git a/curriculum/challenges/english/10-coding-interview-prep/project-euler/problem-72-counting-fractions.md b/curriculum/challenges/english/10-coding-interview-prep/project-euler/problem-72-counting-fractions.md
index ac66afa5a4..3929bde7af 100644
--- a/curriculum/challenges/english/10-coding-interview-prep/project-euler/problem-72-counting-fractions.md
+++ b/curriculum/challenges/english/10-coding-interview-prep/project-euler/problem-72-counting-fractions.md
@@ -8,28 +8,46 @@ dashedName: problem-72-counting-fractions
 
 # --description--
 
-Consider the fraction, `n`/`d`, where n and d are positive integers. If `n`&lt;`d` and HCF(`n`,`d`)=1, it is called a reduced proper fraction.
+Consider the fraction, $\frac{n}{d}$, where `n` and `d` are positive integers. If `n` &lt; `d` and highest common factor, ${HCF}(n, d) = 1$, it is called a reduced proper fraction.
 
 If we list the set of reduced proper fractions for `d` ≤ 8 in ascending order of size, we get:
 
-<div style='text-align: center;'>1/8, 1/7, 1/6, 1/5, 1/4, 2/7, 1/3, 3/8, 2/5, 3/7, 1/2, 4/7, 3/5, 5/8, 2/3, 5/7, 3/4, 4/5, 5/6, 6/7, 7/8</div>
+$$\frac{1}{8}, \frac{1}{7}, \frac{1}{6}, \frac{1}{5}, \frac{1}{4}, \frac{2}{7}, \frac{1}{3}, \frac{3}{8}, \frac{2}{5}, \frac{3}{7}, \frac{1}{2}, \frac{4}{7}, \frac{3}{5}, \frac{5}{8}, \frac{2}{3}, \frac{5}{7}, \frac{3}{4}, \frac{4}{5}, \frac{5}{6}, \frac{6}{7}, \frac{7}{8}$$
 
-It can be seen that there are 21 elements in this set.
+It can be seen that there are `21` elements in this set.
 
-How many elements would be contained in the set of reduced proper fractions for `d` ≤ 1,000,000?
+How many elements would be contained in the set of reduced proper fractions for `d` ≤ `limit`?
 
 # --hints--
 
-`countingFractions()` should return a number.
+`countingFractions(8)` should return a number.
 
 ```js
-assert(typeof countingFractions() === 'number');
+assert(typeof countingFractions(8) === 'number');
 ```
 
-`countingFractions()` should return 303963552391.
+`countingFractions(8)` should return `21`.
 
 ```js
-assert.strictEqual(countingFractions(), 303963552391);
+assert.strictEqual(countingFractions(8), 21);
+```
+
+`countingFractions(20000)` should return `121590395`.
+
+```js
+assert.strictEqual(countingFractions(20000), 121590395);
+```
+
+`countingFractions(500000)` should return `75991039675`.
+
+```js
+assert.strictEqual(countingFractions(500000), 75991039675);
+```
+
+`countingFractions(1000000)` should return `303963552391`.
+
+```js
+assert.strictEqual(countingFractions(1000000), 303963552391);
 ```
 
 # --seed--
@@ -37,16 +55,36 @@ assert.strictEqual(countingFractions(), 303963552391);
 ## --seed-contents--
 
 ```js
-function countingFractions() {
+function countingFractions(limit) {
 
   return true;
 }
 
-countingFractions();
+countingFractions(8);
 ```
 
 # --solutions--
 
 ```js
-// solution required
+function countingFractions(limit) {
+  const phi = {};
+  let count = 0;
+
+  for (let i = 2; i <= limit; i++) {
+    if (!phi[i]) {
+      phi[i] = i;
+    }
+    if (phi[i] === i) {
+      for (let j = i; j <= limit; j += i) {
+        if (!phi[j]) {
+          phi[j] = j;
+        }
+        phi[j] = (phi[j] / i) * (i - 1);
+      }
+    }
+    count += phi[i];
+  }
+
+  return count;
+}
 ```