diff --git a/src/Python/Problem038.py b/src/Python/Problem038.py
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+++ b/src/Python/Problem038.py
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+#!/usr/bin/env python3
+"""
+Created on 03 Jun 2021
+
+@author: David Doblas Jiménez
+@email: daviddoji@pm.me
+
+Solution for problem 38 of Project Euler
+https://projecteuler.net/problem=38
+"""
+
+from utils import timeit
+
+
+@timeit("Problem 38")
+def compute():
+    """
+    Take the number 192 and multiply it by each of 1, 2, and 3:
+
+      192 × 1 = 192
+      192 × 2 = 384
+      192 × 3 = 576
+
+    By concatenating each product we get the 1 to 9 pandigital, 
+    192384576. We will call 192384576 the concatenated product of 
+    192 and (1,2,3)
+
+    The same can be achieved by starting with 9 and multiplying by 
+    1, 2, 3, 4, and 5, giving the pandigital, 918273645, which is 
+    the concatenated product of 9 and (1,2,3,4,5).
+
+    What is the largest 1 to 9 pandigital 9-digit number that can
+    be formed as the concatenated product of an integer with 
+    (1,2, ... , n) where n > 1?
+
+    """
+
+    results = []
+    # Number must 4 digits (exactly) to be pandigital
+    # if n > 1
+    for i in range(1, 10_000):
+        integer = 1
+        number = ""
+        while len(number) < 9:
+            number += str(integer * i)
+            if sorted(number) == list('123456789'):
+                results.append(number)
+    
+            integer += 1
+    
+    return max(results)
+
+
+if __name__ == "__main__":
+
+    print(f"Result for Problem 38: {compute()}")