Solution to problem 48

src/Python/Problem048.py

@@ -0,0 +1,28 @@
+#!/usr/bin/env python3
+"""
+Created on 12 Sep 2021
+
+@author: David Doblas Jiménez
+@email: daviddoji@pm.me
+
+Solution for problem 48 of Project Euler
+https://projecteuler.net/problem=48
+"""
+
+from utils import timeit
+
+
+@timeit("Problem 48")
+def compute():
+    """
+    The series, 1^1 + 2^2 + 3^3 + ... + 10^10 = 10405071317.
+
+    Find the last ten digits of the series, 1^1 + 2^2 + 3^3 + ... + 1000^1000.
+    """
+    series = sum(i**i for i in range(1,1001))
+    return str(series)[-10:]
+
+
+if __name__ == "__main__":
+
+    print(f"Result for Problem 48: {compute()}")