diff --git a/curriculum/challenges/english/10-coding-interview-prep/project-euler/problem-88-product-sum-numbers.md b/curriculum/challenges/english/10-coding-interview-prep/project-euler/problem-88-product-sum-numbers.md
index 96efafdf5d..3134d9cacf 100644
--- a/curriculum/challenges/english/10-coding-interview-prep/project-euler/problem-88-product-sum-numbers.md
+++ b/curriculum/challenges/english/10-coding-interview-prep/project-euler/problem-88-product-sum-numbers.md
@@ -8,7 +8,7 @@ dashedName: problem-88-product-sum-numbers
 
 # --description--
 
-A natural number, N, that can be written as the sum and product of a given set of at least two natural numbers, {`a`<sub>1</sub>, `a`<sub>2</sub>, ... , `a`<sub>k</sub>} is called a product-sum number: N = `a`<sub>1</sub> + `a`<sub>2</sub> + ... + `a`<sub>k</sub> = `a`<sub>1</sub> × `a`<sub>2</sub> × ... × `a`<sub>k</sub>.
+A natural number, `N`, that can be written as the sum and product of a given set of at least two natural numbers, $\\{a_1, a_2, \ldots , a_k\\}$ is called a product-sum number: $N = a_1 + a_2 + \cdots + a_k = a_1 × a_2 × \cdots × a_k$.
 
 For example, 6 = 1 + 2 + 3 = 1 × 2 × 3.
 
@@ -20,26 +20,50 @@ For a given set of size, `k`, we shall call the smallest N with this property a
   <var>k</var>=4: 8 = 1 × 1 × 2 × 4 = 1 + 1 + 2 + 4<br>
   <var>k</var>=5: 8 = 1 × 1 × 2 × 2 × 2  = 1 + 1 + 2 + 2 + 2<br>
   <var>k</var>=6: 12 = 1 × 1 × 1 × 1 × 2 × 6 = 1 + 1 + 1 + 1 + 2 + 6
-</div>
+</div><br>
 
-Hence for 2≤`k`≤6, the sum of all the minimal product-sum numbers is 4+6+8+12 = 30; note that 8 is only counted once in the sum.
+Hence for 2 ≤ `k` ≤ 6, the sum of all the minimal product-sum numbers is 4 + 6 + 8 + 12 = 30; note that `8` is only counted once in the sum.
 
-In fact, as the complete set of minimal product-sum numbers for 2≤`k`≤12 is {4, 6, 8, 12, 15, 16}, the sum is 61.
+In fact, as the complete set of minimal product-sum numbers for 2 ≤ `k` ≤ 12 is $\\{4, 6, 8, 12, 15, 16\\}$, the sum is `61`.
 
-What is the sum of all the minimal product-sum numbers for 2≤`k`≤12000?
+What is the sum of all the minimal product-sum numbers for 2 ≤ `k` ≤ `limit`?
 
 # --hints--
 
-`productSumNumbers()` should return a number.
+`productSumNumbers(6)` should return a number.
 
 ```js
-assert(typeof productSumNumbers() === 'number');
+assert(typeof productSumNumbers(6) === 'number');
 ```
 
-`productSumNumbers()` should return 7587457.
+`productSumNumbers(6)` should return `30`.
 
 ```js
-assert.strictEqual(productSumNumbers(), 7587457);
+assert.strictEqual(productSumNumbers(6), 30);
+```
+
+`productSumNumbers(12)` should return `61`.
+
+```js
+assert.strictEqual(productSumNumbers(12), 61);
+```
+
+`productSumNumbers(300)` should return `12686`.
+
+```js
+assert.strictEqual(productSumNumbers(300), 12686);
+```
+
+`productSumNumbers(6000)` should return `2125990`.
+
+```js
+assert.strictEqual(productSumNumbers(6000), 2125990);
+```
+
+`productSumNumbers(12000)` should return `7587457`.
+
+```js
+assert.strictEqual(productSumNumbers(12000), 7587457);
 ```
 
 # --seed--
@@ -47,16 +71,40 @@ assert.strictEqual(productSumNumbers(), 7587457);
 ## --seed-contents--
 
 ```js
-function productSumNumbers() {
+function productSumNumbers(limit) {
 
   return true;
 }
 
-productSumNumbers();
+productSumNumbers(6);
 ```
 
 # --solutions--
 
 ```js
-// solution required
+function productSumNumbers(limit) {
+  function getProductSums(curProduct, curSum, factorsCount, start) {
+    const k = curProduct - curSum + factorsCount;
+    if (k <= limit) {
+      if (curProduct < minimalProductSums[k]) {
+        minimalProductSums[k] = curProduct;
+      }
+      for (let i = start; i < Math.floor((limit / curProduct) * 2) + 1; i++) {
+        getProductSums(curProduct * i, curSum + i, factorsCount + 1, i);
+      }
+    }
+  }
+
+  const minimalProductSums = new Array(limit + 1).fill(2 * limit);
+  getProductSums(1, 1, 1, 2);
+
+  const uniqueProductSums = [...new Set(minimalProductSums.slice(2))];
+
+  let sum = 0;
+  for (let i = 0; i < uniqueProductSums.length; i++) {
+    sum += uniqueProductSums[i];
+  }
+
+  return sum;
+}
 ```