Smallest number whose sum of digits is square of N

Given an integer N, the task is to find the smallest number whose sum of digits is N².

Examples:

Input: N = 4 
Output: 79 
2⁴ = 16 
sum of digits of 79 = 76

Input: N = 6 
Output: 9999 
2¹⁰ = 1024 which has 4 digits 

Approach: The idea is to find the general term for the smallest number whose sum of digits is square of N. That is 

// First Few terms 
First Term = 1 // N = 1
Second Term = 4 // N = 2
Third Term = 9 // N = 3
Fourth Term = 79 // N = 4
.
.
Nth Term:

*** QuickLaTeX cannot compile formula:
 

*** Error message:
Error: Nothing to show, formula is empty

 <sup>(n^2 \% 9 + 1) * 10 ^ {\lfloor n^2/9 \rfloor} - 1 </sup>

Below is the implementation of the above approach:

C++

// C++ implementation of the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to return smallest
// number whose sum of digits is n^2
int smallestNum(int n)
{
    return (n * n % 9 + 1) * pow(10, n * n / 9) - 1;
}

// Driver Code
int main()
{
    int n = 4;
    cout << smallestNum(n);

    return 0;
}

Java

// Java implementation of the above approach
import java.util.*;

class GFG{

// Function to return smallest
// number whose sum of digits is n^2
static int smallestNum(int n)
{
    return (int)((n * n % 9 + 1) *
         Math.pow(10, n * n / 9) - 1);
}

// Driver Code
public static void main(String[] args)
{
    int n = 4;

    System.out.print(smallestNum(n));
}
}

// This code is contributed by Rajput-Ji

Python 3

# Python implementation of the above approach

# Function to return smallest
# number whose sum of digits is n^2
def smallestNum(n):

    return ((n * n % 9 + 1)  *
          pow(10, int(n * n / 9)) - 1)

# Driver Code

# Given N
N = 4

print(smallestNum(N))

# This code is contributed by Vishal Maurya.

C#

// C# implementation of the above approach
using System;

class GFG{

// Function to return smallest
// number whose sum of digits is n^2
static int smallestNum(int n)
{
    return (int)((n * n % 9 + 1) *
         Math.Pow(10, n * n / 9) - 1);
}

// Driver Code
public static void Main(String[] args)
{
    int n = 4;

    Console.Write(smallestNum(n));
}
}

// This code is contributed by Rajput-Ji

Javascript

<script>
// Javascript implementation of the above approach

    // Function to return smallest
    // number whose sum of digits is n^2
    function smallestNum( n) {
        return parseInt (n * n % 9 + 1) * Math.pow(10, parseInt(n*n / 9)) -1;
    }

    // Driver Code

        let n = 4;

        document.write(smallestNum(n));


// This code contributed by aashish1995
</script>

Output

10
79

Time complexity: O(log₁₀n²)  for given n, as pow function is being used
Auxiliary space: O(1)