Automorphic Number (original) (raw)

Last Updated : 20 Jan, 2025

An **automorphic number is a number whose square ends in the same digits as the number itself. A number n is called **automorphic if:

**n 2 ****(mod 10** d ) = n

where d is the number of digits in n.

**Examples:

**n = 5:
- n2 = 25
- Last digit of 25 is 5, so 5 is automorphic.
**n = 76:
- n2 = 5776
- Last two digits of 5776 are 76, so 76 is automorphic.
**n = 6:
- n2 = 36
- Last digit of 36 is 6, so 6 is automorphic.
**n = 25:
- n2 = 625
- Last two digits of 625 are 25, so 25 is automorphic.

Given a number N, the task is to check whether the number is an Automorphic number or not. A number is called an Automorphic number if and only if its square ends in the same digits as the number itself.

**Examples :

Input : N = 76
Output : Automorphic
Explanation: As 76*76 = 5776

Input : N = 25
Output : Automorphic
As 25*25 = 625**

**Input : N = 7
**Output : Not Automorphic
As 7*7 = 49

Try It Yourself

**Approach:

Store the square of given number.
Loop until N becomes 0 as we have to match all digits with its square.
- Check if (n%10 == sq%10) i.e. last digit of number = last digit of square or not
  * if not equal, return false.
  * Otherwise, continue i.e. reduce the number and square i.e. n = n/10 and sq = sq/10;
Return true if all digits matched.

Below is the implementation of the above approach:

C++ `

// C++ program to check if a number is Automorphic #include using namespace std;

// Function to check Automorphic number bool isAutomorphic(int N) {

  if(N < 0) N = -N;
  

  // Store the square
int sq = N * N;

// Start Comparing digits
while (N > 0) {
    // Return false, if any digit of N doesn't
    // match with its square's digits from last
    if (N % 10 != sq % 10)
        return false;

    // Reduce N and square
    N /= 10;
    sq /= 10;
}

return true;

}

// Driver code int main() { int N = 5;

isAutomorphic(N) ? cout << "Automorphic"
                 : cout << "Not Automorphic";

return 0;

}

Java

// Java program to check if a number is Automorphic import java.io.*; class Test { // Function to check Automorphic number static boolean isAutomorphic(int N) { // Store the square if(N < 0) N = -N; int sq = N * N;

    // Start Comparing digits
    while (N > 0) {
        // Return false, if any digit of N doesn't
        // match with its square's digits from last
        if (N % 10 != sq % 10)
            return false;

        // Reduce N and square
        N /= 10;
        sq /= 10;
    }

    return true;
}

// Driver method
public static void main(String[] args)
{
    int N = 5;

    System.out.println(isAutomorphic(N) ? "Automorphic" : "Not Automorphic");
}

}

Python

Python program to check if a number is Automorphic

Function to check Automorphic number

def isAutomorphic(N):

# Store the square
if N < 0:
  N = -N
sq = N * N
 
# Start Comparing digits
while (N > 0) :

    # Return false, if any digit of N doesn't
    # match with its square's digits from last
    if (N % 10 != sq % 10) :
        return False

    # Reduce N and square
    N //= 10
    sq //= 10

return True

Driver code

N = 5 if isAutomorphic(N) : print ("Automorphic") else : print ("Not Automorphic")

This Code is contributed by Nikita Tiwari.

C#

// C# program to check if a // number is Automorphic using System;

class GFG {

// Function to check Automorphic number
static bool isAutomorphic(int N)
{

    // Store the square
      if(N < 0) N = -N;
    int sq = N * N;

    // Start Comparing digits
    while (N > 0) {
        // Return false, if any digit
        // of N doesn't match with its
        // square's digits from last
        if (N % 10 != sq % 10)
            return false;

        // Reduce N and square
        N /= 10;
        sq /= 10;
    }

    return true;
}

// Driver Code
public static void Main()
{
    int N = 5;

    Console.Write(isAutomorphic(N) ? "Automorphic" : "Not Automorphic");
}

}

// This code is Contributed by Nitin Mittal.

JavaScript

PHP

N=−N = -N=−N; sq=sq = sq=N * $N; // Start Comparing digits while ($N > 0) { // Return false, if any // digit of N doesn't // match with its square's // digits from last if ($N % 10 != $sq % 10) return -1; // Reduce N and square $N /= 10; $sq /= 10; } return 1; } // Driver code $N = 5; geeks=isAutomorphic(geeks = isAutomorphic(geeks=isAutomorphic(N) ? "Automorphic" : "Not Automorphic"; echo $geeks; // This code is contributed by ajit ?>

**Time Complexity: O(log10N)
**Auxiliary Space: O(1)

**Another Approach to Solve the Problem

Do first check if number is negative then make it positive.
Store the square of number.
Find the count of the digit of the number sothat you can find the count of digit of last number of the square of the number equal to the number i.e it doesn't mean if the count of digit of last number of square is equal to the number will be equal to each other.
And after counting the digit of the number perform : - **squareNum%power(10, count)
Finally check the last number of square of number is equal to number or not.

_Let's see the implementation as explained for above approach : -

C++ `

#include #include <math.h> using namespace std; bool checkAuto(int a){ if(a < 0) a = -a; int squareNum = a*a; int temp = a; int count = 0; // count of digit of a int lastNum = 0; while(temp > 0){ count++; temp = temp/10; } int lastDigit = (squareNum)%(int(pow(10, count))); if(lastDigit == a) return true; else return false; } int main() { int num = -4; if(checkAuto(num)) cout << "Automorphic"; else cout << "Not Automorphic"; cout << endl; return 0; }

Java

import java.io.; import java.util.; public class Solution { public static void main(String[] args) { Scanner sc = new Scanner(System.in); int a = -4; if(a < 0) a = -a; int squareNum = a*a; int temp = a; int count = 0; // count of digit of a while(temp > 0){ count++; temp = temp/10; } int lastDigit = squareNum%(int)Math.pow(10, count); // System.out.print(lastDigit); if(lastDigit == a) System.out.print("Automorphic"); else System.out.print("Not Automorphic");

}

Python

def checkAuto(a): if a < 0: a = -a squareNum = a*a temp = a count = 0 while temp != 0: count += 1 temp = int(temp/10) lastDigit = squareNum%pow(10, count) if lastDigit == a: return "Automorphic" else: return "Not Automorphic" num = -4 print(checkAuto(num))

C#

using System;

class Solution { static void Main(string[] args) { int a = -4; if (a < 0) a = -a; int squareNum = a * a; int temp = a; int count = 0; // count of digit of a while (temp > 0) { count++; temp = temp / 10; } int lastDigit = squareNum % (int)Math.Pow(10, count); // Console.Write(lastDigit); if (lastDigit == a) Console.Write("Automorphic"); else Console.Write("Not Automorphic"); } }

JavaScript

function checkAuto(a){ if(a < 0) a = -a; let squareNum = a*a; let temp = a; let count = 0; // count of digit of a while(temp > 0){ count++; temp = Math.floor(temp/10); } let lastDigit = (squareNum)%(Math.pow(10, count)); if(lastDigit == a) return 1; else return 0; } let num = -4; if(checkAuto(num)) console.log("Automorphic"); else console.log("Not Automorphic");

**Time Complexity: - O(log10N), where N is the given number.
**Auxiliary Space:- O(1)