diff --git a/Competitive_Programming/C++/Maths & Number Theory/Euler_toitent_function.ipynb b/Competitive_Programming/C++/Maths & Number Theory/Euler_toitent_function.ipynb
new file mode 100644
index 0000000000..d9d3a99b4b
--- /dev/null
+++ b/Competitive_Programming/C++/Maths & Number Theory/Euler_toitent_function.ipynb	
@@ -0,0 +1,308 @@
+{
+  "nbformat": 4,
+  "nbformat_minor": 0,
+  "metadata": {
+    "colab": {
+      "name": "Euler_toitent_function.md",
+      "provenance": [],
+      "collapsed_sections": []
+    },
+    "kernelspec": {
+      "name": "python3",
+      "display_name": "Python 3"
+    },
+    "language_info": {
+      "name": "python"
+    }
+  },
+  "cells": [
+    {
+      "cell_type": "markdown",
+      "metadata": {
+        "id": "dP9LH9ODdmir"
+      },
+      "source": [
+        "# **EULER TOITENT FUNCTION:**"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "metadata": {
+        "id": "4WWVz2rf8MLD"
+      },
+      "source": [
+        "**PROBLEM STATEMENT:**\n",
+        "\n",
+        "*   The Objective is to explain the euler totient function algorithm,analyse its complexity and working ,and state its uses in computation\n",
+        "\n",
+        "\n",
+        "\n"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "metadata": {
+        "id": "UCesKC4Jd6yA"
+      },
+      "source": [
+        "Euler’s Totient function Φ (n) for an input n is the count of numbers belonging to set {1,2,3,....n} such that the number is coprime with n, i.e gcd(m,n)=1 such that 1<=m<n.\n",
+        "\n",
+        "\n",
+        "e.g :\n",
+        "\n",
+        "> Φ(1) = 1  \n",
+        "gcd(1, 1) is 1\n",
+        "\n",
+        "> Φ(2) = 1\n",
+        "gcd(1, 2) is 1, but gcd(2, 2) is 2.\n",
+        "\n",
+        "\n",
+        ">Φ(3) = 2\n",
+        "gcd(1, 3) is 1 and gcd(2, 3) is 1\n",
+        "\n",
+        "> Φ(4) = 2\n",
+        "gcd(1, 4) is 1 and gcd(3, 4) is 1\n",
+        "\n",
+        "\n",
+        "> Φ(5) = 4\n",
+        "gcd(1, 5) is 1, gcd(2, 5) is 1, \n",
+        "gcd(3, 5) is 1 and gcd(4, 5) is 1\n",
+        "\n",
+        "\n",
+        "> Φ(6) = 2\n",
+        "gcd(1, 6) is 1 and gcd(5, 6) is 1,\n",
+        "\n",
+        "\n"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "metadata": {
+        "id": "45B_hNXcefjl"
+      },
+      "source": [
+        "**PROPERTIES:**\n",
+        "\n",
+        "\n",
+        "1.   Φ (n)= n*(1-1/p1)*(1-1/p2)....*(1-1/pk),  where p1,p2 .....,pk are prime factors of n.\n",
+        "\n",
+        "2.   Φ (p)=p-1, where p=prime number\n",
+        "\n",
+        "3.   Φ (ab)=Φ (a)*Φ (b)\n",
+        "\n",
+        "4.   Φ (p^k)=(p^k)-(p^(k-1)) ,where p=prime number\n",
+        "\n",
+        "\n",
+        "\n"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "metadata": {
+        "id": "XquPBAxYgZ3z"
+      },
+      "source": [
+        "**PROOF:**\n",
+        "\n",
+        "> Let n=(p1^a)*(p2^b)......*(pk^k) ,where p1,p2 .....,pk are prime factors of n and a,b,....k are powers of prime factors of n.\n",
+        "\n",
+        "> Then, Φ (n)=Φ ((p1^a)*(p2^b)......*(pk^k))\n",
+        "\n",
+        "> Using The Properties we get: Φ (n)= Φ ((p1^a))*Φ ((p2^b))*.....Φ ((pk^k)),\n",
+        "\n",
+        "\n",
+        "> Since,p1,p2,...pk are prime numbers Therefore,Using the Properties Φ (pi^x)=(pi^x)-(pi^(x-1))\n",
+        "\n",
+        "\n",
+        "> => Φ (pi^x) = (pi^x)*(1-1/pi)\n",
+        "\n",
+        "> => Φ (n)= ((p1^a)*(1-1/p1))*((p2^b)*(1-1/p2))*........((pk^k)*(1-1/pk))\n",
+        "\n",
+        "\n",
+        "> => Φ (n) = ((p1^a)*(p2^b)......*(pk^k))*(1-1/p1)*(1-1/p2)*.....(1-1/pk)\n",
+        "\n",
+        "\n",
+        "> Therefore, Φ (n) = n*(1-1/p1)*(1-1/p2)*.....(1-1/pk)\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "metadata": {
+        "id": "j9zS8ZH8-cNt"
+      },
+      "source": [
+        "**ALGORITHM:**\n",
+        "\n",
+        "\n",
+        "1.   CREATE AN ARRAY PHI OF SIZE N+1 AND INITIALIZE iTh ELEMENT WITH VALUE i ,i={1,2,.....N}\n",
+        "2.   RUN A LOOP FROM 2 TO N, AND IF ELEMENT IN ARRAY PHI IS EQUAL TO ITS INDEX(WHICH MEANS ELEMENT IS PRIME),CHANGE ITS VALUE TO INDEX-1\n",
+        "3. FOR ALL MULTIPLES OF THE CURRENT INDEX LESS THAN EQUAL TO N,MULTIPLY THE ELEMENTS WITH THE VALUE= (1-1/MULTIPLE).\n",
+        "4. RETURN VALUE AT PHI[N]\n",
+        "\n",
+        "\n",
+        "\n"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "metadata": {
+        "id": "cc--D9uuj10W"
+      },
+      "source": [
+        "**FUNCTION TO CALCULATE EULER TOTIENT FUNCTION FOR A NATURAL NUMBER N:(IN C++)**"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "metadata": {
+        "id": "Ku3SKXSEkFZN"
+      },
+      "source": [
+        "int euler_totient_function(int n){\n",
+        "    int phi[n+1];\n",
+        "    for(int i=1;i<=n;i++){\n",
+        "        phi[i]=i;\n",
+        "    }\n",
+        "    for(int i=2;i<=n;i++){\n",
+        "        if(phi[i]==i){\n",
+        "            phi[i]=i-1;\n",
+        "            for(int j=2*i;j<=n;j+=i){\n",
+        "                phi[j]=(phi[j]*(i-1))/i;\n",
+        "            }\n",
+        "        }\n",
+        "    }\n",
+        "    return phi[n];\n",
+        "}"
+      ],
+      "execution_count": null,
+      "outputs": []
+    },
+    {
+      "cell_type": "markdown",
+      "metadata": {
+        "id": "3cTCqyagBPoX"
+      },
+      "source": [
+        "**COMPLEXITY ANALYSIS:**\n",
+        "\n",
+        "\n",
+        "1.   SPACE COMPLEXITY:\n",
+        "SINCE,WE ARE USING ONLY AN EXTRA ARRAY OF SIZE N+1,SO OUR SPACE COMPLEXITY IS **O(n)**\n",
+        "\n",
+        "2.   TIME COMPLEXITY:\n",
+        "Since,We start our iteration from i=2 and go till the last multiple of 2 less than equal to n .\n",
+        "Therefore,our time complexity would be N/2 + N/3 +N/5 +N/7+.....N/p;where p is the last prime less than equal to n\n",
+        "This Sum comes out to be Nlog(log(N))\n",
+        "Thus,Our Time Complexity is **O(Nlog(log(N)))**"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "metadata": {
+        "id": "8nnAF_Q7HaFi"
+      },
+      "source": [
+        "**SOLVING A PROBLEM BASED ON EULER TOTIENT FUNCTION:**\n",
+        "\n",
+        "PROBLEM STATEMENT:\n",
+        "Given n, calculate and print the sum :\n",
+        "LCM(1,n) + LCM(2,n) + .. + LCM(n,n)\n",
+        "where LCM(i,n) denotes the Least Common Multiple of the integers i and n.\n",
+        "\n",
+        "ALGORITHM:\n",
+        "USING LCM,GCD AND EULER'S TOTIENT FUNCTION PROPERTIES WE ARRIVE AT THE LCM SUM AS, **LCM SUM = n/2*(SUM(Φ (d)*d)+1) ,Where d=all divisors of n except 1**"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "metadata": {
+        "id": "hZGlLO17IMKz"
+      },
+      "source": [
+        "void euler(long long n,long long *phi){\n",
+        "    for(int i=1;i<=n;i++){\n",
+        "        phi[i]=i;\n",
+        "    }\n",
+        "    for(int i=2;i<=n;i++){\n",
+        "        if(phi[i]==i){\n",
+        "            phi[i]=i-1;\n",
+        "            for(int j=2*i;j<=n;j+=i){\n",
+        "                phi[j]=(phi[j]*(i-1))/i;\n",
+        "            }\n",
+        "        }\n",
+        "    }\n",
+        "}\n",
+        "void func(long long n)\n",
+        "{\n",
+        "    long long *phi=new long long[n+1];\n",
+        "    euler(n,phi);\n",
+        "    long long sum=0;\n",
+        "    for(int i=1;i<=n;i++){\n",
+        "        if(n%i==0){\n",
+        "            sum+=(phi[i]*i);\n",
+        "        }\n",
+        "    }\n",
+        "    sum=((sum+1)*n)/2;\n",
+        "    cout<<sum<<endl;\n",
+        "    delete[] phi;\n",
+        "}"
+      ],
+      "execution_count": null,
+      "outputs": []
+    },
+    {
+      "cell_type": "markdown",
+      "metadata": {
+        "id": "H-uYnESAEdVo"
+      },
+      "source": [
+        "**APPLICATION:**\n",
+        "\n",
+        "1. **The RSA cryptosystem:**\n",
+        "\n",
+        "* Setting up an RSA system involves choosing large prime numbers p and q, computing n = pq and k = φ(n), and finding two numbers e and d such that ed ≡ 1 (mod k). The numbers n and e (the \"encryption key\") are released to the public, and d (the \"decryption key\") is kept private.\n",
+        "\n",
+        "* A message, represented by an integer m, where 0 < m < n, is encrypted by computing S = me (mod n).\n",
+        "\n",
+        "* It is decrypted by computing t = Sd (mod n). Euler's Theorem can be used to show that if 0 < t < n, then t = m.\n",
+        "\n",
+        "* The security of an RSA system would be compromised if the number n could be factored or if φ(n) could be computed without factoring n.\n",
+        "\n"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "metadata": {
+        "id": "p7poM8KdG4P6"
+      },
+      "source": [
+        "**REFERENCES:**\n",
+        "\n",
+        "\n",
+        "*   wikipedia/euler's totient function\n",
+        "*   cp-algorithms/euler's totient function\n",
+        "*   geeksforgeeks/euler's totient function\n",
+        "\n"
+      ]
+    }
+  ]
+}