From 2b5b2c6304f68112484a599af6ce6121088e14c2 Mon Sep 17 00:00:00 2001
From: Ravi Kandasamy Sundaram <ksr1122@gmail.com>
Date: Sat, 10 Oct 2020 19:59:02 +0200
Subject: [PATCH] Added solution for Project Euler problem 119 (#2931)

Name: Digit power sum

Problem Statement: The number 512 is interesting because it is equal to the sum of its digits raised to some power: 5 + 1 + 2 = 8, and 83 = 512. Another example of a number with this property is 614656 = 284.
We shall define an to be the nth term of this sequence and insist that a number must contain at least two digits to have a sum.
You are given that a2 = 512 and a10 = 614656.
Find a30

Reference: https://projecteuler.net/problem=119

reference: #2695

Co-authored-by: Ravi Kandasamy Sundaram <rkandasamysundaram@luxoft.com>
---
 project_euler/problem_119/__init__.py |  0
 project_euler/problem_119/sol1.py     | 51 +++++++++++++++++++++++++++
 2 files changed, 51 insertions(+)
 create mode 100644 project_euler/problem_119/__init__.py
 create mode 100644 project_euler/problem_119/sol1.py

diff --git a/project_euler/problem_119/__init__.py b/project_euler/problem_119/__init__.py
new file mode 100644
index 000000000..e69de29bb
diff --git a/project_euler/problem_119/sol1.py b/project_euler/problem_119/sol1.py
new file mode 100644
index 000000000..7f343ac24
--- /dev/null
+++ b/project_euler/problem_119/sol1.py
@@ -0,0 +1,51 @@
+"""
+Problem 119: https://projecteuler.net/problem=119
+
+Name: Digit power sum
+
+The number 512 is interesting because it is equal to the sum of its digits
+raised to some power: 5 + 1 + 2 = 8, and 8^3 = 512. Another example of a number
+with this property is 614656 = 28^4. We shall define an to be the nth term of
+this sequence and insist that a number must contain at least two digits to have a sum.
+You are given that a2 = 512 and a10 = 614656. Find a30
+"""
+
+import math
+
+
+def digit_sum(n: int) -> int:
+    """
+    Returns the sum of the digits of the number.
+    >>> digit_sum(123)
+    6
+    >>> digit_sum(456)
+    15
+    >>> digit_sum(78910)
+    25
+    """
+    return sum([int(digit) for digit in str(n)])
+
+
+def solution(n: int = 30) -> int:
+    """
+    Returns the value of 30th digit power sum.
+    >>> solution(2)
+    512
+    >>> solution(5)
+    5832
+    >>> solution(10)
+    614656
+    """
+    digit_to_powers = []
+    for digit in range(2, 100):
+        for power in range(2, 100):
+            number = int(math.pow(digit, power))
+            if digit == digit_sum(number):
+                digit_to_powers.append(number)
+
+    digit_to_powers.sort()
+    return digit_to_powers[n - 1]
+
+
+if __name__ == "__main__":
+    print(solution())