Java: Check whether a number is an Armstrong Number or not
Java Numbers: Exercise-28 with Solution
Write a Java program to check whether a number is an Armstrong Number or not.
Armstrong (Michael F. Armstrong) number is a number that is equal to the sum of cubes of its digits. For example 0, 1, 153, 370, 371 and 407 are the Armstrong numbers
Sample Solution:
Java Code:
import java.util.*;
public class solution {
public static boolean is_Amstrong(int n) {
int remainder, sum = 0, temp = 0;
temp = n;
while (n > 0) {
remainder = n % 10;
sum = sum + (remainder * remainder * remainder);
n = n / 10;
}
return sum == temp;
}
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
System.out.print("Input an integer: ");
String input = scanner.nextLine();
int number = Integer.parseInt(input);
System.out.println("Is Armstrong number? "+is_Amstrong(number));
}
}
Sample Output:
Input an integer: 153 Is Armstrong number? true
Flowchart:

Java Code Editor:
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Previous: Write a Java program to create the first twenty Hamming numbers.
Next: Write a Program in Java to check whether a number is a Lucky Number or not.
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Java: Tips of the Day
IsPowerOfTwo
Checks if a value is positive power of two.
To understand how it works let's assume we made a call IsPowerOfTwo(4).
As value is greater than 0, so right side of the && operator will be evaluated.
The result of (~value + 1) is equal to value itself. ~100 + 001 => 011 + 001 => 100. This is equal to value.
The result of (value & value) is value. 100 & 100 => 100.
This will value the expression to true as value is equal to value.
public static boolean isPowerOfTwo(final int value) { return value > 0 && ((value & (~value + 1)) == value); }
Ref: https://bit.ly/3sA5d4I
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