A very simple program to list down the Armstrong numbers in Java.
I am currently accepting programming and web development work kindly contact me in the following email address for further details. Thank you.
My mobile number here in the Philippines is 09173084360.
public class Main {
public static void main(String[] args) {
int a =0;
int num =0;
int n, sum, temp, remainder, digits;
int start = 100;
int end =2000;
System.out.println("\n\n");
System.out.print("List of Armstrong Number in Java");
System.out.println("\n");
for (int i = start; i <= end; i++) {
sum = 0;
digits = 0;
temp = i;
while (temp != 0) {
digits++;
temp = temp / 10;
}
temp = i;
while (temp != 0) {
remainder = temp % 10;
sum = sum + power(remainder, digits);
temp = temp / 10;
}
if (i == sum)
System.out.println(i + " is an Armstrong number.");
}
System.out.println();
System.out.print("End of Program");
System.out.println();
}
static int power(int n, int r) {
int c, p = 1;
for (c = 1; c <= r; c++)
p = p * n;
return p;
}
}
I am currently accepting programming and web development work kindly contact me in the following email address for further details. Thank you.
My email address are the following jakerpomperada@gmail.com and jakerpomperada@yahoo.com.
My telephone number at home here in Bacolod City, Negros Occidental Philippines is +63 (034) 4335675.
Sample Program Output
Program Listing
public class Main {
public static void main(String[] args) {
int a =0;
int num =0;
int n, sum, temp, remainder, digits;
int start = 100;
int end =2000;
System.out.println("\n\n");
System.out.print("List of Armstrong Number in Java");
System.out.println("\n");
for (int i = start; i <= end; i++) {
sum = 0;
digits = 0;
temp = i;
while (temp != 0) {
digits++;
temp = temp / 10;
}
temp = i;
while (temp != 0) {
remainder = temp % 10;
sum = sum + power(remainder, digits);
temp = temp / 10;
}
if (i == sum)
System.out.println(i + " is an Armstrong number.");
}
System.out.println();
System.out.print("End of Program");
System.out.println();
}
static int power(int n, int r) {
int c, p = 1;
for (c = 1; c <= r; c++)
p = p * n;
return p;
}
}
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