[Hacker Rank]: Project Euler #1: Multiples of 3 and 5 solved ✓

Author: Gonzalo Diaz

src/hackerrank/projecteuler/euler001.md

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+# [Multiples of 3 and 5](https://www.hackerrank.com/contests/projecteuler/challenges/euler001)
+
+- Difficulty: #easy
+- Category: #ProjectEuler+
+
+If we list all the natural numbers below $ 10 $ that are multiples of
+ $ 3 $ or $ 5 $, we get $ 3, 5, 6 $ and $ 9 $.
+ The sum of these multiples
+ is $ 23 $.
+
+Find the sum of all the multiples of $ 3 $ or $ 5 $ below $ N $.
+
+## Input Format
+
+First line contains $ T $ that denotes the number of test cases.
+This is followed by $ T $ lines, each containing an integer, $ N $.
+
+## Constraints
+
+- 1 $ \leq T \leq 10^5 $
+- 1 $ \leq N \leq 10^9 $
+
+## Output Format
+
+For each test case, print an integer that denotes the sum of all the multiples
+of $ 3 $ or $ 5 $ below $ N $.
+
+## Sample Input 0
+
+```text
+2
+10
+100
+```
+
+## Sample Output 0
+
+```text
+23
+2318
+```
+
+## Explanation 0
+
+For $ N = 10 $, if we list all the natural numbers below $ 10 $ that are
+ multiples of $ 3 $ or $ 5 $, we get $ 3, 5, 6 $ and $ 9 $.
+  The sum of these multiples is $ 23 $.
+
+Similarly for $ N = 100 $, we get $ 2318 $.

src/hackerrank/projecteuler/euler001.py

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+# pylint: disable=C0103:invalid-name
+
+import math
+
+
+# Function to find sum of Arithmetic Progression series
+def sum_ap(n: int, d: int):
+
+    # Number of terms
+    n = n//d
+
+    return (n) * (1 + n) * d // 2
+
+
+# Function to find the sum of all multiples of a and b below n
+def euler001(a, b, n):
+
+    # Since, we need the sum of multiples less than N
+    n = n-1
+    lcm = (a*b)//math.gcd(a, b)
+
+    return sum_ap(n, a) + \
+        sum_ap(n, b) - \
+        sum_ap(n, lcm)

src/hackerrank/projecteuler/euler001_test.py

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+import unittest
+from .euler001 import euler001
+
+
+class TestEuler001(unittest.TestCase):
+
+    def test_euler001(self):
+
+        tests = [
+          {'a': 3, 'b': 5, 'n': 10, 'answer': 23},
+          {'a': 3, 'b': 5, 'n': 100, 'answer': 2318},
+          {'a': 3, 'b': 5, 'n': 1000, 'answer': 233168}
+        ]
+
+        for _, _tt in enumerate(tests):
+
+            self.assertEqual(
+                euler001(_tt['a'], _tt['b'], _tt['n']), _tt['answer'],
+                f"{_} | euler001({_tt['a']}, {_tt['b']}, {_tt['n']}) must be "
+                f"=> {_tt['answer']}")