diff --git a/math/modular_inverse_fermat_little_theorem.cpp b/math/modular_inverse_fermat_little_theorem.cpp
new file mode 100644
index 000000000..965c21298
--- /dev/null
+++ b/math/modular_inverse_fermat_little_theorem.cpp
@@ -0,0 +1,79 @@
+/*
+ * C++ Program to find the modular inverse using Fermat's Little Theorem.
+ * Fermat's Little Theorem state that => ϕ(m) = m-1, where m is a prime number.
+ *
+ * (a * x) ≡ 1 mod m.
+ * x ≡ (a^(-1)) mod m.
+ *
+ * Using Euler's theorem we can modify the equation.
+ *
+ * (a^ϕ(m)) ≡ 1 mod m (Where '^' denotes the exponent operator)
+ * Here 'ϕ' is Euler's Totient Function. For modular inverse existence 'a' and 'm' must be relatively primes numbers.
+ * To apply Fermat's Little Theorem is necessary that 'm' must be a prime number.
+ * Generally in many competitive programming competitions 'm' is either 1000000007 (1e9+7) or 998244353.
+ *
+ * We considered m as large prime (1e9+7).
+ * (a^ϕ(m)) ≡ 1 mod m (Using Euler's Theorem)
+ * ϕ(m) = m-1 using Fermat's Little Theorem.
+ * (a^(m-1)) ≡ 1 mod m
+ * Now multiplying both side by (a^(-1)).
+ * (a^(m-1)) * (a^(-1)) ≡  (a^(-1)) mod m
+ * (a^(m-2)) ≡  (a^(-1)) mod m
+ *
+ * We will find the exponent using binary exponentiation. Such that the algorithm works in O(log(m)) time.
+ *
+ * Example: -
+ * a = 3 and m = 7
+ * (a^(-1) mod m) is equivalent to (a^(m-2) mod m)
+ * (3^(5) mod 7) = (243 mod 7) = 5
+ * Hence, ( 3^(-1) mod 7 ) = 5
+ * or ( 3 * 5 ) mod 7 = 1 mod 7 (as a*(a^(-1)) = 1)
+ */
+
+#include<iostream>
+#include<vector>
+
+// Recursive function to calculate exponent in O(log(n)) using binary exponent.
+int64_t binExpo(int64_t a, int64_t b, int64_t m) {
+    a %= m;
+    int64_t res = 1;
+    while (b > 0) {
+        if (b%2) {
+            res = res * a % m;
+        }
+        a = a * a % m;
+        // Dividing b by 2 is similar to right shift.
+        b >>= 1;
+    }
+    return res;
+}
+
+// Prime check in O(sqrt(m)) time.
+bool isPrime(int64_t m) {
+    if (m <= 1) {
+        return false;
+    } else {
+        for (int i=2; i*i <= m; i++) {
+            if (m%i == 0) {
+                return false;
+            }
+        }
+    }
+    return true;
+}
+
+int main() {
+    int64_t a, m;
+    // Take input of  a and m.
+    std::cout << "Computing ((a^(-1))%(m)) using Fermat's Little Theorem";
+    std::cout << std::endl << std::endl;
+    std::cout << "Give input 'a' and 'm' space separated : ";
+    std::cin >> a >> m;
+    if (isPrime(m)) {
+        std::cout << "The modular inverse of a with mod m is (a^(m-2)) : ";
+        std::cout << binExpo(a, m-2, m) << std::endl;
+    } else {
+        std::cout << "m must be a prime number.";
+        std::cout << std::endl;
+    }
+}