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 5Input: 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++ 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;
} |
// Java program to print the digital // roots of a number import java.io.*;
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 |
# Python3 program to print the digital # roots of a number # Function to return dig-sum def summ(n):
if (n = = 0 ):
return 0 ;
if (n % 9 = = 0 ):
return 9 ;
else :
return (n % 9 );
# Function to print the Digital Roots def printDigitalRoot(n):
# store the largest digital roots
maxi = 1 ;
dig = 1 ;
# Iterate till sqrt(n)
for i in range ( 1 , int ( pow (n, 1 / 2 ) + 1 )):
# if i is a factor
if (n % i = = 0 ):
# get the digit sum of both
# factors i and n/i
d1 = summ(n / i);
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
print ( int (dig), " " , int (maxi));
# Driver Code if __name__ = = '__main__' :
n = 10 ;
# Function call to prdigital roots
printDigitalRoot(n);
# This code is contributed by 29AjayKumar |
// 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 |
<script> // javascript program to print the digital // roots of a number // Function to return dig-sum function summ(n) {
if (n == 0)
return 0;
return (n % 9 == 0) ? 9 : (n % 9);
}
// Function to print the Digital Roots
function printDigitalRoot(n) {
// store the largest digital roots
var maxi = 1;
var dig = 1;
// Iterate till sqrt(n)
for (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
var d1 = summ(n / i);
var 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
document.write(dig + " " + maxi);
}
// Driver Code
var n = 10;
// Function call to print digital roots
printDigitalRoot(n);
// This code is contributed by todaysgaurav </script> |
5 5
Time Complexity: O(sqrt(N))
Auxiliary Space: O(1)