feat: Check if a number is an Armstrong number (3-digit)

🧠 Logic:
- Take integer input `n`, store original in `m`.
- Initialize `sum = 0`.
- Extract digits using `% 10` and compute cube of each digit.
- Accumulate sum of cubes.
- After loop, compare `sum` with original number `m`.
  - If equal → Armstrong number.

Signed-off-by: Somesh diwan <[email protected]>

Author: Someshdiwan

Section8LoopAB/src/ArmsStrongNumber.java

@@ -1,14 +1,13 @@
 /*
 Find a number is armstrong number or not.
-A number is armsstrong, sum of the cubs of digits of number, n =153,
-3 cube + 5 cube + 1 cube + 153 then its Armsstrong number.
+A number is arms strong, sum of the cubs of digits of number, n =153,
+3 cubes + 5 cubes + 1 cube = 153 then its Armstrong number.
 */
 
 import java.util.Scanner;
 
 public class ArmsStrongNumber {
-    public static void main(String[] args)
-    {
+    public static void main(String[] args) {
         Scanner scan = new Scanner(System.in);
         System.out.println("Enter a number: ");
         int n= scan.nextInt();
@@ -17,11 +16,9 @@ public static void main(String[] args)
         int sum=0;
         int r;
 
-        while(n>0)
-        {
+        while(n>0) {
             r=n%10;
             n=n/10;
-
             sum=sum+r*r*r;
         }
         if(sum==m)