Programming in java to find the no. is niven or not.. by scanner..
siddhartharao77:
Is it even?
Answers
Answered by
1
import java.util.*;
public class Niven
{
public static void main(String args[])
{
Scanner demo = new Scanner(System.in);
System.out.print("Enter some number ");
int a = demo.nextInt();
int b = a, d, e = 0;
while(b>0)
{
d = b%10;
e = e + d;
b = b/10;
}
if(a%e == 0)
System.out.println("It is Niven Number.");
else
System.out.println("Not Niven Number");
}
}
Hope this helps!
public class Niven
{
public static void main(String args[])
{
Scanner demo = new Scanner(System.in);
System.out.print("Enter some number ");
int a = demo.nextInt();
int b = a, d, e = 0;
while(b>0)
{
d = b%10;
e = e + d;
b = b/10;
}
if(a%e == 0)
System.out.println("It is Niven Number.");
else
System.out.println("Not Niven Number");
}
}
Hope this helps!
Answered by
1
import java.util.*;
class niven_number
{
public void main()
{
Scanner sc=new Scanner(System.in);
System.out.println("Enter a number to check whether it is a niven number or not");
int n=sc.nextInt();
int d=0;
int s=0;
int cpy=n;
while(n>0)
{
d=n%10;
s=s+d;
n=n/10;
}
if(cpy%s==0)
{
System.out.println(cpy+" is a niven number");
}
else
{
System.out.println(cpy+" is not a niven number");
}
}
}
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