From c5276b46f3bf33848cbf0ab3fae22fa76803482a Mon Sep 17 00:00:00 2001
From: utkarsh <ucponia@gmail.com>
Date: Tue, 6 Oct 2020 22:41:44 +0530
Subject: [PATCH 1/3] Added Problem 72

---
 project_euler/problem_72/__init__.py |  0
 project_euler/problem_72/sol1.py     | 46 ++++++++++++++++++++++++++++
 2 files changed, 46 insertions(+)
 create mode 100644 project_euler/problem_72/__init__.py
 create mode 100644 project_euler/problem_72/sol1.py

diff --git a/project_euler/problem_72/__init__.py b/project_euler/problem_72/__init__.py
new file mode 100644
index 000000000000..e69de29bb2d1
diff --git a/project_euler/problem_72/sol1.py b/project_euler/problem_72/sol1.py
new file mode 100644
index 000000000000..ce6268a0d1a4
--- /dev/null
+++ b/project_euler/problem_72/sol1.py
@@ -0,0 +1,46 @@
+"""
+Problem 72 Counting fractions: https://projecteuler.net/problem=72
+
+Description:
+
+Consider the fraction, n/d, where n and d are positive integers. If n<d and HCF(n,d)=1,
+it is called a reduced proper fraction.
+If we list the set of reduced proper fractions for d ≤ 8 in ascending order of size, we
+get: 1/8, 1/7, 1/6, 1/5, 1/4, 2/7, 1/3, 3/8, 2/5, 3/7, 1/2, 4/7, 3/5, 5/8, 2/3, 5/7,
+3/4, 4/5, 5/6, 6/7, 7/8
+It can be seen that there are 21 elements in this set.
+How many elements would be contained in the set of reduced proper fractions for
+d ≤ 1,000,000?
+
+Solution:
+
+Number of numbers between 1 and n that are coprime to n is given by the Euler's Totient
+function, phi(n). So, the answer is simply the sum of phi(n) for 2 <= n <= 1,000,000
+Sum of phi(d), for all d|n = n. This result can be used to find phi(n) using a sieve.
+
+Time: 3.5 sec
+"""
+
+
+def solution(N: int = 1_000_000) -> int:
+    """
+    Returns an integer, the solution to the problem
+    >>> solution(10)
+    31
+    >>> solution(100)
+    3043
+    >>> solution(1_000_000)
+    303963552391
+    """
+
+    phi = [i - 1 for i in range(N + 1)]
+
+    for i in range(2, N + 1):
+        for j in range(2 * i, N + 1, i):
+            phi[j] -= phi[i]
+
+    return sum(phi[2 : N + 1])
+
+
+if __name__ == "__main__":
+    print(solution(100))

From 50059e5614cdf1722b400339a4695fa6e97b2f70 Mon Sep 17 00:00:00 2001
From: utkarsh <ucponia@gmail.com>
Date: Tue, 6 Oct 2020 22:47:34 +0530
Subject: [PATCH 2/3] Removed args from solution()

---
 project_euler/problem_72/sol1.py | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/project_euler/problem_72/sol1.py b/project_euler/problem_72/sol1.py
index ce6268a0d1a4..03ac1813935b 100644
--- a/project_euler/problem_72/sol1.py
+++ b/project_euler/problem_72/sol1.py
@@ -43,4 +43,4 @@ def solution(N: int = 1_000_000) -> int:
 
 
 if __name__ == "__main__":
-    print(solution(100))
+    print(solution())

From cacd8b082ef3d276130550cbddf5750cf7d80dcb Mon Sep 17 00:00:00 2001
From: utkarsh <ucponia@gmail.com>
Date: Mon, 12 Oct 2020 17:30:29 +0530
Subject: [PATCH 3/3] Incorporated the suggested changes

---
 project_euler/problem_72/sol1.py | 14 +++++++-------
 1 file changed, 7 insertions(+), 7 deletions(-)

diff --git a/project_euler/problem_72/sol1.py b/project_euler/problem_72/sol1.py
index 03ac1813935b..846396ab0f9c 100644
--- a/project_euler/problem_72/sol1.py
+++ b/project_euler/problem_72/sol1.py
@@ -22,24 +22,24 @@
 """
 
 
-def solution(N: int = 1_000_000) -> int:
+def solution(limit: int = 1_000_000) -> int:
     """
     Returns an integer, the solution to the problem
     >>> solution(10)
     31
     >>> solution(100)
     3043
-    >>> solution(1_000_000)
-    303963552391
+    >>> solution(1_000)
+    304191
     """
 
-    phi = [i - 1 for i in range(N + 1)]
+    phi = [i - 1 for i in range(limit + 1)]
 
-    for i in range(2, N + 1):
-        for j in range(2 * i, N + 1, i):
+    for i in range(2, limit + 1):
+        for j in range(2 * i, limit + 1, i):
             phi[j] -= phi[i]
 
-    return sum(phi[2 : N + 1])
+    return sum(phi[2 : limit + 1])
 
 
 if __name__ == "__main__":