TheAlgorithms · raklaptudirm · Oct 8, 2021 · Oct 5, 2021 · Oct 5, 2021 · Oct 5, 2021
@@ -190,6 +190,7 @@
   * [Problem013](https://github.com/TheAlgorithms/Javascript/blob/master/Project-Euler/Problem013.js)
   * [Problem014](https://github.com/TheAlgorithms/Javascript/blob/master/Project-Euler/Problem014.js)
   * [Problem015](https://github.com/TheAlgorithms/Javascript/blob/master/Project-Euler/Problem015.js)
+  * [Problem016](https://github.com/TheAlgorithms/Javascript/blob/master/Project-Euler/Problem016.js)
   * [Problem020](https://github.com/TheAlgorithms/Javascript/blob/master/Project-Euler/Problem020.js)
   * [Problem1](https://github.com/TheAlgorithms/Javascript/blob/master/Project-Euler/Problem1.js)
   * [Problem10](https://github.com/TheAlgorithms/Javascript/blob/master/Project-Euler/Problem10.js)

@@ -0,0 +1,43 @@
+/**
+ * Problem 16 - Power digit sum
+ *
+ * @see {@link https://projecteuler.net/problem=16}
+ *
+ * 2¹⁵ = 32768 and the sum of its digits is 3 + 2 + 7 + 6 + 8 = 26.
+ *
+ * What is the sum of the digits of the number 2¹⁰⁰⁰ ?
+ */
+
+/**
+ * Returns the power digit sum of n^pow.
+ *
+ * @param {number} [n=2]
+ * @param {number} [pow=1000]
+ * @returns {number}
+ */
+const powerDigitSum = function (n = 2, pow = 1000) {
+  // The idea is to consider each digit (d*10^exp) separately, right-to-left.
+  // digits = [units, tens, ...]
+
+  const digits = [n]
+  let p = 1
+
+  while (++p <= pow) {
+    let carry = 0
+    for (let exp = 0; exp < digits.length; exp++) {
+      const prod = digits[exp] * n + carry
+      carry = Math.floor(prod / 10)
+      digits[exp] = prod % 10
+    }
+    while (carry > 0) {
+      digits.push(carry % 10)
+      carry = Math.floor(carry / 10)
+    }
+  }
+
+  // (digits are reversed but we only want the sum so it doesn't matter)
+
+  return digits.reduce((prev, current) => prev + current, 0)
+}
+
+export { powerDigitSum }
@@ -0,0 +1,16 @@
+import { powerDigitSum } from '../Problem016'
+
+describe('Check Problem 16 - Power digit sum', () => {
+  it('Power digit sum of 2^15', () => {
+    expect(powerDigitSum(2, 15)).toBe(26)
+  })
+
+  it('Power digit sum of 2^1000', () => {
+    expect(powerDigitSum()).toBe(1366)
+    expect(powerDigitSum(2, 1000)).toBe(1366)
+  })
+
+  it('Power digit sum of 3^5000', () => {
+    expect(powerDigitSum(3, 5000)).toBe(11097)
+  })
+})