Determine if a number is an Armstrong number

[chezkaty]

can you please help me again to this?:( cause the previous one I've asked to you i didn't get it :((

write a program that determines whether a given positive integer is an armstrong number.

-----armstrong number is an n-digit number that is equal to the sum of the nTH power of each digit.

your program should contain the following:

A. funtion that returns a value when x is raised to y.
B. funtions that determines the number of digits in an n-digit number.

use these funtions to determine if a number is an armstrong number.

thankyou. much appreciated.

[Anirudh Rb]

The standard library function pow(x, y) returns the result when x is raised to y.

The following function returns the number of digits in a number n.

int getNOOfDigits(int n)
{
   int count = 0;

   while(n > 0)
   {
      count++;
      n /= 10; //Divide an integer by 10 and see what happens.
   }
   return count;
}

Using these functions it becomes very easy to check if a number is Armstrong.

Note that you can obtain the last digit of a number by:

r = n % 10; //here r is the last digit of n.

Now you can work the rest out yourself!!!
Believe me, Its real easy. Anyone can work it out!!!

[ashok1514]

#include<stdio.h>
#include<conio.h>

int main()
{
    int n,i,rev=0,num;
    printf("\n Enter a number : ");
    scanf("%d",&n);
    num=n;
    while(n) {
        rev=rev*10+n%10;
        n/=10;
    }
    if(num==rev)
        printf("\n Is Armstrong");
    else
        printf("\n Not an Armstrong");
}

[Anirudh Rb]

#include<stdio.h>
#include<conio.h>

int main()
{
    int n,i,rev=0,num;
    printf("\n Enter a number : ");
    scanf("%d",&n);
    num=n;
    while(n) {
        rev=rev*10+n%10;
        n/=10;
    }
    if(num==rev)
        printf("\n Is Armstrong");
    else
        printf("\n Not an Armstrong");
}

This program does not tell whether a number is armstrong or not. This is a program to test whether a number is equal to its reverse...