# Java – Calculating Digital Root, is there a better way

javamath

This is how i calculated the digital root of an integer.

``````import acm.program.*;

public class Problem7 extends ConsoleProgram
{
public void run()
{
println("This program calculates the digital root of an interger.");

int num = readInt("Enter the number: ");
int sum = 0;
while (true)
{
if (num > 0)
{
int dsum = num % 10;
num /= 10;
sum += dsum;
}
else if (sum > 9)
{
int dsum = sum % 10;
sum /= 10;
sum += dsum;

} else if (sum <= 9 ) break;
}
println("Digital Root is: " + sum);
}
``````

The program works fine.

Is there a better/shorter way of calculating the digital root of a number. ?

EDIT/ADDED : Here is the implementation of the above problem by using Tyler's answer, it works as well:

``````import acm.program.*;

public class Problem7 extends ConsoleProgram
{
public void run()
{
println("This program calculates the digital root of an interger.");

int num = readInt("Enter the number: ");
println("Digital Root of " + num + " is: " + (1 + (num - 1) % 9));
}
}
``````

#### Best Solution

``````#include <stdio.h>

int main(void)
{
int number;
scanf("%d", &number);

printf("The digital root of %d is %d.", number, (1 + (number - 1) % 9));
}
``````

Had I not been able to find Ramans' formula this is how I would write this program...:

``````#include <stdio.h>
#include <ctype.h>

int main(void)
{
int c;
int number = 0;
while ((c = getchar()) != EOF)
{
if (isdigit(c))
number += (c - '0');
}
if (number <= 9)
{
printf("The digital root is %d\n", number);
}
else
{
printf("%d", number);
}

}
``````

After compiling, to run this, basically you just chain these together. I believe four is the most you could possibly need for an integer.

``````\$ echo 829382938 | ./digitalroot | ./digitalroot | ./digitalroot | ./digitalroot
``````