C program and algorithm to check whether a number is armstrong or not.

ALGORITHM

1. Start
2. read number
3. set sum=0 and duplicate=number
4. reminder=number%10
5. sum=sum+(reminder*reminder*reminder)
6. number=number/10
7. repeat steps 4 to 6 until number > 0
8. if sum = duplicate
9. display number is armstrong
10. else 
11. display number is not armstrong
12. stop

PROGRAM

#include<stdio.h>

#include<conio.h>

void main()

{

clrscr();

int n,s,r,dup;

s=0;

printf("\n Enter a number please:\t");

scanf("%d",&n);

dup=n;

while(n>0)

{

r=n%10;

s=s+(r*r*r);       //or you can use pow(r,3) and use the header file math.h

n=n/10;

}

if(dup=s)

printf("\n %d is an armstrong number",dup);

else

printf("\n %d is not an armstrong

number",dup);

getch();

}

LOGIC

An armstrong number: the sum of cube of its digits results in that number itself.

eg: 153 = (1*1*1)+(5*5*5)+(3*3*3) = 1+125+27 =153.

r=n%10:

153%10 gives the reminder, that is the number 3 here. The modulus operator is used here to seperate the digits. The other thing is that " a small number % large number= that small number itself ".

eg: 3%5= 3 (reminder)

n=n/10:

after a modulus operation we will get the last digit of that number now we have to remove the last digit from the input number so we divide it by 10 and what happens is:

153%10=3 (reminder,r=3)

sum=0+(3*3*3) (sum,s=27)

153/10=15 (number,n=15)

the actual answer is 15.3 but since we initialize it as an integer the floating point value isn't considered. So the calculation continues like:

now the value of n is 15 after the division by 10.

15%10=5                      (reminder,r=5)

sum=(3*3*3)+(5*5*5)  (sum,s=152)

15/10=1                         (number,n=1)

and this continues till the number becomes 0 that is the while condition (n>0) evaluates to false.

OUTPUT

Enter a number please: 153

153 is an armstrong number