How to check if a number is Armstrong in Java
In order to understand if a number is an Armstrong number in Python, one should be well versed with the following topics:
- Conditional statements
- Loops
What is an Armstrong number?
A positive integer is called an Armstrong number of order n if it is equal to the sum of the powers of its own digits. For example:
abc…(up to n digits) = a^n + b^n +c^n ...(up to n digits)
In the case of the Armstrong number of 3 digits, it is equal to the sum of the cubes for each digit. For example, the number 407 = 4 x 4 x 4 + 0 x 0 x 0 + 7 x 7 x 7 ⇒ 407 is an Armstrong number.
Code
Let’s take a look at the code below.
import java.util.Scanner;import java.lang.Math;class CheckArmstrongNumber{static boolean isArmstrong(int n){int temp, digits=0, last=0, sum=0;temp = n;while(temp>0){temp = temp/10;digits++;}temp = n;while(temp>0){last = temp % 10;sum += (Math.pow(last, digits));temp = temp/10;}if(n == sum)return true;elsereturn false;}public static void main(String args[]){int num;Scanner sc= new Scanner(System.in);num = sc.nextInt();if(isArmstrong(num))System.out.print(num + " is Armstrong.");elseSystem.out.print(num + " is not Armstrong.");}}
Enter the input below
Explanation
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In lines 1 and 2, we import the required header files.
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In lines 5 to 25, we create a function in which we run a loop to find the sum of the power of the order of each digit. We obtain the unit digit each time with the modulus operator
%, since the remainder of the number when it is divided by 10 is the last digit of that number. -
In lines 15 to 20, we take the power of the order of the number with the
pow()function. -
In lines 29 to 31, we read the input number from the user.
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In lines 33 to 37, we call the
isArmstrong()function to check whether the number is Armstrong and print the statement accordingly.
