Check whether a number is Tech Number or not

Last Updated : 23 Mar, 2024

Given a number, the task is to check if it is a Tech number or not.

Note: A Tech number is a number that has an even number of digits and if the number is split into two equal halves(From the middle), then the square of the sum of these halves is equal to the number itself.

Examples:

Input: n = 81
Output: True
Explanation: 8 + 1 = 9 and 9²= 81

Input: n = 2025
Output: True
Explanation: 20 + 25 = 45 and 45²= 2025

Input: n = 1521
Output: False
Explanation: 15 + 21 = 36 and 36²= 1296 which is not equal to 1521

Approach: Let us understand how to check whether a number is a tech number or not.

Check if the number of digits is even. If it’s not even, the number cannot be a Tech number.
Split the number into two equal halves and fin
Compare the squared sum with the original number. If they are equal, then the number is a tech number; otherwise, it’s not.

Below is the implementation of the above approach:

Java

// Java program to check if a
// number is Tech number or not
public class TechNumber {
    // Function to count the
    // number of digits in a number
    static int countDigits(int n)
    {
        int count = 0;
        while (n > 0) {
            n /= 10;
            count++;
        }
        return count;
    }

    // Returns true if n is a tech
    // number, else false
    static boolean isTech(int n)
    {
        // Count the number of digits
        int count = countDigits(n);

        // Return false if digits are odd
        if (count % 2 != 0)
            return false;

        // Split the number in two
        // equal halve and store their sum
        int half = count / 2;
        int first_half = n / (int)Math.pow(10, half);
        int second_half = n % (int)Math.pow(10, half);
        int sum = first_half + second_half;

        // Check if the square of the sum
        // is equal to the original number
        if ((sum * sum) == n)
            return true;
        else
            return false;
    }

    // Driver code
    public static void main(String[] args)
    {
        if (isTech(81))
            System.out.println("True");
        else
            System.out.println("False");

        if (isTech(2025))
            System.out.println("True");
        else
            System.out.println("False");

        if (isTech(1521))
            System.out.println("True");
        else
            System.out.println("False");
    }
}
// This code is contributed by
// Abhishek Kumar

using System;

namespace TechNumberExample
{
    class Program
    {
        // Function to count the number of digits in a number
        static int CountDigits(int n)
        {
            int count = 0;
            while (n > 0)
            {
                n /= 10;
                count++;
            }
            return count;
        }

        // Function to check if a number is a Tech number
        static bool IsTech(int n)
        {
            int count = CountDigits(n);

            // Return false if the number of digits is odd
            if (count % 2 != 0)
                return false;

            int half = count / 2;
            int firstHalf = n / (int)Math.Pow(10, half);
            int secondHalf = n % (int)Math.Pow(10, half);
            int sum = firstHalf + secondHalf;

            // Check if the square of the sum is equal to the original number
            if ((sum * sum) == n)
                return true;
            else
                return false;
        }

        static void Main(string[] args)
        {
            // Test cases
            if (IsTech(81))
                Console.WriteLine("True");
            else
                Console.WriteLine("False");

            if (IsTech(2025))
                Console.WriteLine("True");
            else
                Console.WriteLine("False");

            if (IsTech(1521))
                Console.WriteLine("True");
            else
                Console.WriteLine("False");
        }
    }
}

Javascript

// Javascript program to check if 
//a number is Tech number or not
// Function to count the number 
//of digits in a number
function countDigits(n)
{
    let count = 0;
    while (n > 0)
    {
        n = Math.floor(n / 10);
        count++;
    }
    return count;
}

// Returns true if n is a tech 
//number, else false
function isTech(n)
{
    // Count the number of digits
    const count = countDigits(n);

    // Return false if digits are odd
    if (count % 2 !== 0)
        return false;

    // Split the number in two equal halves
    //and store their sum
    const half = Math.floor(count / 2);
    const first_half = Math.floor(n / 10 ** half);
    const second_half = n % 10 ** half;
    const sum = first_half + second_half;

    // Check if the square of the sum
    //is equal to the original number
    if (sum * sum === n)
        return true;
    else
        return false;
}

// Driver code
if (isTech(81)) 
    console.log("True");
else
    console.log("False");

if (isTech(2025))
    console.log("True");
else
    console.log("False");

if (isTech(1521))
    console.log("True");
else
    console.log("False");
// This code is contributed by
// Abhishek Kumar

C++14

// C++ program to check if a
// number is Tech number or not
#include <bits/stdc++.h>
using namespace std;

// Function to count the number
// of digits in a number
int countDigits(int n)
{
    int count = 0;
    while (n > 0) {
        n /= 10;
        count++;
    }
    return count;
}

// Returns true if n is a tech number, else false
bool istech(int n)
{

    // Count the number of digits
    int count = countDigits(n);

    // Return false if digits are odd
    if (count % 2 != 0)
        return false;

    // Split the number in two equal
    // halves and store their sum
    int half = count / 2;
    int first_half = n / pow(10, half);
    int second_half = n % int(pow(10, half));
    int sum = first_half + second_half;

    // Check if the square of the sum
    // is equal to the original number
    if ((sum * sum) == n)
        return true;
    else
        return false;
}

// Driver code
int main()
{
    if (istech(81))
        cout << "True" << endl;
    else
        cout << "False" << endl;
    if (istech(2025))
        cout << "True" << endl;
    else
        cout << "False" << endl;
    if (istech(1521))
        cout << "True" << endl;
    else
        cout << "False" << endl;
    return 0;
}

Python3

import math

# Function to count the number of digits in a number
def count_digits(n):
    count = 0
    while n > 0:
        n //= 10
        count += 1
    return count

# Returns true if n is a tech number, else false
def is_tech(n):
    # Count the number of digits
    count = count_digits(n)

    # Return false if digits are odd
    if count % 2 != 0:
        return False

    # Split the number in two equal halves and store their sum
    half = count // 2
    first_half = n // 10 ** half
    second_half = n % (10 ** half)
    sum_val = first_half + second_half

    # Check if the square of the sum is equal to the original number
    if sum_val ** 2 == n:
        return True
    else:
        return False

# Driver code
if is_tech(81):
    print("True")
else:
    print("False")
if is_tech(2025):
    print("True")
else:
    print("False")
if is_tech(1521):
    print("True")
else:
    print("False")