diff --git a/math/number_of_positive_divisors.cpp b/math/number_of_positive_divisors.cpp
new file mode 100644
index 000000000..48ab63c36
--- /dev/null
+++ b/math/number_of_positive_divisors.cpp
@@ -0,0 +1,66 @@
+/// C++ Program to calculate number of divisors.
+
+#include<iostream>
+#include<vector>
+
+/**
+ * This algorithm use the prime factorization approach.
+ * Any number can be written in multiplication of its prime factors.
+ * Let N = P1^E1 * P2^E2 ... Pk^Ek
+ * Therefore. number-of-divisors(N) = (E1+1) * (E2+1) ... (Ek+1).
+ * Where P1, P2 ... Pk are prime factors and E1, E2 ... Ek are exponents respectively.
+ *
+ * Example:-
+ * N = 36
+ * 36 = (3^2 * 2^2)
+ * number_of_positive_divisors(36) = (2+1) * (2+1) = 9.
+ * list of positive divisors of 36 = 1, 2, 3, 4, 6, 9, 12, 18, 36.
+ *
+ * Similarly if N is -36 at that time number of positive divisors remain same.
+ *
+ * Example:-
+ * N = -36
+ * -36 = -1 * (3^2 * 2^2)
+ * number_of_positive_divisors(-36) = (2+1) * (2+1) = 9.
+ * list of positive divisors of -36 = 1, 2, 3, 4, 6, 9, 12, 18, 36.
+ *
+**/
+
+int number_of_positive_divisors(int n) {
+    std::vector<int> prime_exponent_count;
+    for (int i=2; i*i <= n; i++) {
+        int prime_count = 0;
+        while (n % i == 0) {
+            prime_count += 1;
+            n /= i;
+        }
+        if (prime_count != 0) {
+            prime_exponent_count.push_back(prime_count);
+        }
+    }
+    if (n > 1) {
+        prime_exponent_count.push_back(1);
+    }
+
+    int divisors_count = 1;
+
+    for (int i=0; i < prime_exponent_count.size(); i++) {
+        divisors_count = divisors_count * (prime_exponent_count[i]+1);
+    }
+
+    return divisors_count;
+}
+
+int main() {
+    int n;
+    std::cin >> n;
+    if (n < 0) {
+        n = -n;
+    }
+    if (n == 0) {
+        std::cout << "All non-zero numbers are divisors of 0 !" << std::endl;
+    } else {
+        std::cout << "Number of positive divisors is : ";
+        std::cout << number_of_positive_divisors(n) << std::endl;
+    }
+}