Sudo Placement[1.7] | Greatest Digital Root

Given a number N, you need to find a divisor of N such that the Digital Root of that divisor is the greatest among all other divisors of N. If more than one divisors give the same greatest Digital Root then output the maximum divisor.

The Digital Root of a non-negative number can be obtained by repeatedly summing the digits of the number until we reach to a single digit. Example: DigitalRoot(98)=9+8=>17=>1+7=>8(single digit!). The task is to print the greatest divisor having the greatest Digital Root followed by a space and the Digital Root of that divisor.

Examples:

Input: N = 10
Output: 5 5
The divisors of 10 are: 1, 2, 5, 10. The Digital Roots of these divisors are as follows:
1=>1, 2=>2, 5=>5, 10=>1. The greatest Digital Root is 5 which is produced by divisor 5, so answer is 5 5

Input: N = 18
Output: 18 9
The divisors of 18 are: 1, 2, 3, 6, 9, 18. The Digital Roots of these divisors are as follows:
1=>1, 2=>2, 3=>3, 6=>6, 9=>9, 18=>9. As we can see that 9 and 18 both have the greatest Digital Root of 9. So we select the maximum of those divisors, that is Max(9, 18)=18. So answer is 18 9

A naive approach will be to iterate till N and find all the factors and their digit sums. Store the largest among them and print them.

Time Complexity: O(N)

An efficient approach is to loop till sqrt(N), then the factors will be i and n/i. Check for the largest digit sum among them, in case of similar digit sums, store the larger factor. Once the iteration is completed, print them.

Below is the implementation of the above approach.

C++

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// C++ program to print the
// digital roots of a number
#include <bits/stdc++.h>
using namespace std;
  
// Function to return
// dig-sum
int summ(int n)
{
    if (n == 0)
        return 0;
    return (n % 9 == 0) ? 9 : (n % 9);
}
  
// Function to print the Digital Roots
void printDigitalRoot(int n)
{
  
    // store the largest digital roots
    int maxi = 1;
    int dig = 1;
  
    // Iterate till sqrt(n)
    for (int i = 1; i <= sqrt(n); i++) {
  
        // if i is a factor
        if (n % i == 0) {
            // get the digit sum of both
            // factors i and n/i
            int d1 = summ(n / i);
            int d2 = summ(i);
  
            // if digit sum is greater
            // then previous maximum
            if (d1 > maxi) {
                dig = n / i;
                maxi = d1;
            }
  
            // if digit sum is greater
            // then previous maximum
            if (d2 > maxi) {
                dig = i;
                maxi = d2;
            }
  
            // if digit sum is same as
            // then previous maximum, then
            // check for larger divisor
            if (d1 == maxi) {
                if (dig < (n / i)) {
                    dig = n / i;
                    maxi = d1;
                }
            }
  
            // if digit sum is same as
            // then previous maximum, then
            // check for larger divisor
            if (d2 == maxi) {
                if (dig < i) {
                    dig = i;
                    maxi = d2;
                }
            }
        }
    }
  
    // Print the digital roots
    cout << dig << " " << maxi << endl;
}
  
// Driver Code
int main()
{
    int n = 10;
  
    // Function call to print digital roots
    printDigitalRoot(n);
    return 0;
}

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Java

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// Java program to print the digital
// roots of a number
class GFG
{
      
// Function to return dig-sum
static int summ(int n)
{
    if (n == 0)
        return 0;
    return (n % 9 == 0) ? 9 : (n % 9);
}
  
// Function to print the Digital Roots
static void printDigitalRoot(int n)
{
  
    // store the largest digital roots
    int maxi = 1;
    int dig = 1;
  
    // Iterate till sqrt(n)
    for (int i = 1; i <= Math.sqrt(n); i++) 
    {
  
        // if i is a factor
        if (n % i == 0
        {
            // get the digit sum of both
            // factors i and n/i
            int d1 = summ(n / i);
            int d2 = summ(i);
  
            // if digit sum is greater
            // then previous maximum
            if (d1 > maxi) 
            {
                dig = n / i;
                maxi = d1;
            }
  
            // if digit sum is greater
            // then previous maximum
            if (d2 > maxi) 
            {
                dig = i;
                maxi = d2;
            }
  
            // if digit sum is same as
            // then previous maximum, then
            // check for larger divisor
            if (d1 == maxi)
            {
                if (dig < (n / i))
                {
                    dig = n / i;
                    maxi = d1;
                }
            }
  
            // if digit sum is same as
            // then previous maximum, then
            // check for larger divisor
            if (d2 == maxi) 
            {
                if (dig < i) 
                {
                    dig = i;
                    maxi = d2;
                }
            }
        }
    }
  
    // Print the digital roots
    System.out.println(dig + " " + maxi);
}
  
// Driver Code
public static void main(String[] args)
{
    int n = 10;
  
    // Function call to print digital roots
    printDigitalRoot(n);
}
}
// This code is contributed by mits

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C#

// C# program to print the digital
// roots of a number
using System;

class GFG
{

// Function to return dig-sum
static int summ(int n)
{
if (n == 0)
return 0;
return (n % 9 == 0) ? 9 : (n % 9);
}

// Function to print the Digital Roots
static void printDigitalRoot(int n)
{

// store the largest digital roots
int maxi = 1;
int dig = 1;

// Iterate till sqrt(n)
for (int i = 1; i <= Math.Sqrt(n); i++) { // if i is a factor if (n % i == 0) { // get the digit sum of both // factors i and n/i int d1 = summ(n / i); int d2 = summ(i); // if digit sum is greater // then previous maximum if (d1 > maxi)
{
dig = n / i;
maxi = d1;
}

// if digit sum is greater
// then previous maximum
if (d2 > maxi)
{
dig = i;
maxi = d2;
}

// if digit sum is same as
// then previous maximum, then
// check for larger divisor
if (d1 == maxi)
{
if (dig < (n / i)) { dig = n / i; maxi = d1; } } // if digit sum is same as // then previous maximum, then // check for larger divisor if (d2 == maxi) { if (dig < i) { dig = i; maxi = d2; } } } } // Print the digital roots Console.WriteLine(dig + " " + maxi); } // Driver Code public static void Main() { int n = 10; // Function call to print digital roots printDigitalRoot(n); } } // This code is contributed // by Akanksha Rai [tabbyending]

Output:

5 5

Time Complexity: O(sqrt(N))



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