From 02012c817b9b9aedc8c99513161cc0426d492cd2 Mon Sep 17 00:00:00 2001
From: Tatheer Fathima <148070120+T-Fathima@users.noreply.github.com>
Date: Sun, 10 Nov 2024 03:48:40 +0000
Subject: [PATCH] Added Counting divisors in Number Theory

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+---
+id: counting-divisors
+title: "Counting Divisors"
+sidebar_label: "Number theory"
+sidebar_position: 11
+description: "An algorithm to compute the total number of divisors of a given integer n."
+tags: [Counting Divisors, number theory, factorization, divisors]
+---
+
+# Counting Divisors
+
+The **Counting Divisors** algorithm calculates the total number of divisors of a given integer **n**. Divisors are numbers that divide **n** without leaving a remainder. This algorithm is particularly useful in number theory problems, such as perfect number checking, factorization, and determining the structure of numbers.
+
+## Algorithm to Count Divisors
+
+To count the divisors of a number **n**, iterate through all integers from 1 to √n and check if they divide **n**. If they do, both the divisor and its corresponding pair are counted.
+
+### Steps:
+1. Initialize a counter to 0.
+2. For each integer **i** from 1 to √n:
+   - If **n** is divisible by **i** (i.e., **n % i == 0**), increment the counter.
+   - If **i** is not equal to **n / i**, increment the counter again, as both **i** and **n / i** are divisors.
+3. Return the counter as the total number of divisors.
+
+### Example
+
+For **n = 28**:
+- Divisors: 1, 2, 4, 7, 14, 28.
+- Count of divisors: 6.
+
+### Code Implementation (C++)
+
+```cpp
+#include <iostream>
+#include <cmath>
+using namespace std;
+
+int countDivisors(int n) {
+    int count = 0;
+    for (int i = 1; i <= sqrt(n); i++) {
+        if (n % i == 0) {
+            count++;
+            if (i != n / i) {
+                count++;
+            }
+        }
+    }
+    return count;
+}
+
+int main() {
+    int n = 28;
+    cout << "Number of divisors of " << n << " is: " << countDivisors(n) << endl;
+    return 0;
+}
+```
+
+# Python Implementation
+```python
+import math
+
+def count_divisors(n):
+    count = 0
+    for i in range(1, int(math.sqrt(n)) + 1):
+        if n % i == 0:
+            count += 1
+            if i != n // i:
+                count += 1
+    return count
+
+# Example usage
+n = 28
+print(f"Number of divisors of {n} is: {count_divisors(n)}")
+```
+
+# Time Complexity
+The time complexity of this algorithm is O(√n), as we iterate only up to the square root of n.
+
+# Applications of Counting Divisors
+Counting divisors is widely used in various fields:
+
+1. Perfect Number Checking: To verify if a number is perfect, we need to check if the sum of its divisors equals the number itself.
+2. Prime Factorization: Understanding the structure of numbers by counting and finding the prime divisors.
+3. Mathematical Puzzles: Many problems in competitive programming rely on efficiently counting divisors.
+
+# Key Points to Remember
+Efficient Counting: Instead of iterating up to n, iterate up to √n to reduce time complexity.
+Symmetry of Divisors: For every divisor i, n / i is also a divisor, making the count more efficient.