diff --git a/curriculum/challenges/english/10-coding-interview-prep/project-euler/problem-87-prime-power-triples.md b/curriculum/challenges/english/10-coding-interview-prep/project-euler/problem-87-prime-power-triples.md
index fcf84bac7b..57da5afe79 100644
--- a/curriculum/challenges/english/10-coding-interview-prep/project-euler/problem-87-prime-power-triples.md
+++ b/curriculum/challenges/english/10-coding-interview-prep/project-euler/problem-87-prime-power-triples.md
@@ -8,29 +8,53 @@ dashedName: problem-87-prime-power-triples
 
 # --description--
 
-The smallest number expressible as the sum of a prime square, prime cube, and prime fourth power is 28. In fact, there are exactly four numbers below fifty that can be expressed in such a way:
+The smallest number expressible as the sum of a prime square, prime cube, and prime fourth power is `28`. In fact, there are exactly four numbers below fifty that can be expressed in such a way:
 
 <div style='margin-left: 4em;'>
   28 = 2<sup>2</sup> + 2<sup>3</sup> + 2<sup>4</sup><br>
   33 = 3<sup>2</sup> + 2<sup>3</sup> + 2<sup>4</sup><br>
   49 = 5<sup>2</sup> + 2<sup>3</sup> + 2<sup>4</sup><br>
   47 = 2<sup>2</sup> + 3<sup>3</sup> + 2<sup>4</sup>
-</div>
+</div><br>
 
-How many numbers below fifty million can be expressed as the sum of a prime square, prime cube, and prime fourth power?
+How many numbers below `n` can be expressed as the sum of a prime square, prime cube, and prime fourth power?
 
 # --hints--
 
-`primePowerTriples()` should return a number.
+`primePowerTriples(50)` should return a number.
 
 ```js
-assert(typeof primePowerTriples() === 'number');
+assert(typeof primePowerTriples(50) === 'number');
 ```
 
-`primePowerTriples()` should return 1097343.
+`primePowerTriples(50)` should return `4`.
 
 ```js
-assert.strictEqual(primePowerTriples(), 1097343);
+assert.strictEqual(primePowerTriples(50), 4);
+```
+
+`primePowerTriples(10035)` should return `684`.
+
+```js
+assert.strictEqual(primePowerTriples(10035), 684);
+```
+
+`primePowerTriples(500000)` should return `18899`.
+
+```js
+assert.strictEqual(primePowerTriples(500000), 18899);
+```
+
+`primePowerTriples(5000000)` should return `138932`.
+
+```js
+assert.strictEqual(primePowerTriples(5000000), 138932);
+```
+
+`primePowerTriples(50000000)` should return `1097343`.
+
+```js
+assert.strictEqual(primePowerTriples(50000000), 1097343);
 ```
 
 # --seed--
@@ -38,16 +62,70 @@ assert.strictEqual(primePowerTriples(), 1097343);
 ## --seed-contents--
 
 ```js
-function primePowerTriples() {
+function primePowerTriples(n) {
 
   return true;
 }
 
-primePowerTriples();
+primePowerTriples(50);
 ```
 
 # --solutions--
 
 ```js
-// solution required
+function primePowerTriples(n) {
+  function getSievePrimes(max) {
+    const primes = [];
+    const primesMap = new Array(max).fill(true);
+    primesMap[0] = false;
+    primesMap[1] = false;
+
+    for (let i = 2; i <= max; i += 2) {
+      if (primesMap[i]) {
+        primes.push(i);
+        for (let j = i * i; j <= max; j = j + i) {
+          primesMap[j] = false;
+        }
+      }
+      if (i === 2) {
+        i = 1;
+      }
+    }
+    return primes;
+  }
+
+  function getPowersSummed(numbers, powers, limit, curSum) {
+    if (curSum >= limit) {
+      return [];
+    } else if (powers.length === 0) {
+      return [curSum];
+    }
+
+    const powersSummed = [];
+
+    const curPower = powers[0];
+    const powersLeft = powers.slice(1);
+    for (let i = 0; i < numbers.length; i++) {
+      const curNumber = numbers[i];
+      const nextSum = curSum + curNumber ** curPower;
+      if (nextSum >= limit) {
+        return powersSummed;
+      }
+      const result = getPowersSummed(
+        numbers,
+        powersLeft,
+        limit,
+        curSum + curNumber ** curPower
+      );
+      powersSummed.push(...result);
+    }
+    return powersSummed;
+  }
+
+  const maximumBaseNumber = Math.floor(Math.sqrt(n - 2 ** 3 - 2 ** 4)) + 1;
+  const primes = getSievePrimes(maximumBaseNumber);
+  const uniqueSums = new Set(getPowersSummed(primes, [2, 3, 4], n, 0));
+
+  return uniqueSums.size;
+}
 ```