program to find armstrong number between 1 to 1000

Question

sir why this code is not work i is run only for 1 using System; public class Arm2 { public static void Main() { int reminder i num; for(i=1; i { int sum = 0; num = i; while(num>0) { reminder = num % 10; sum = reminder*reminder*reminder; num = num/10; } if(sum == i) { Console.WriteLine({0} is armstrong number. i); } } Console.ReadLine(); } }

Vulpes · Answer

Can I point out that there's also a fundamental problem here with the definition of an Armstrong number. Technically an Armstrong number is an N digit number which is the sum of its digits raised to the power N . See for example this link: http://everything2.net/index.pl?node_id=1407017&displaytype=printable&lastnode_id=1407017 So: for one digit numbers you raise to the power 1. for two digit numbers you raise to the power 2. for three digit numbers you raise to the power 3. In other words you only cube the digits if it's a 3 digit number. Allowing for this the program should now read: using System; public class Arm2 { public static void Main() { int remainder i num; for (i = 1; i { int sum = 0; num = i; while (num > 0) { remainder = num % 10; if (i { sum += remainder; } else if (i { sum += remainder * remainder; } else { sum += remainder * remainder * remainder; } num = num / 10; } if (sum == i) { Console.WriteLine({0 -3} is armstrong number. i); } } Console.ReadLine(); } } and the output is: 1 is armstrong number. 2 is armstrong number. 3 is armstrong number. 4 is armstrong number. 5 is armstrong number. 6 is armstrong number. 7 is armstrong number. 8 is armstrong number. 9 is armstrong number. 153 is armstrong number. 370 is armstrong number. 371 is armstrong number. 407 is armstrong number.

Tom Mohan · Answer

That was a small mistake instead of sum = reminder*reminder*reminder; use sum += reminder * reminder * reminder; pleas see the changes. public static void Main() { int reminder i num; for (i = 1; i { int sum = 0; num = i; while (num > 0) { reminder = num % 10; sum += reminder * reminder * reminder; num = num / 10; } if (sum == i) { Console.WriteLine({0} is armstrong number. i); } } Console.ReadLine(); } } Mark it as Accepted Answer If its useful