To find the given number is armstrong
(or) not
Program:
#include<stdio.h>
#include<conio.h>
void main()
{
int n, r, s = 0, t;
clrscr();
printf("Enter a number : ");
scanf("%d", &n);
t = n;
while(n > 0)
{
r = n % 10;
s = s + (r * r * r);
n = n / 10;
}
if(s == t)
printf("\n%d is an armstrong number", t);
else
printf("\n%d is not an armstrong number", t);
getch();
}
Output:
Case: 1
Enter a number : 153
153 is an armstrong number
Case: 2
Enter a number : 123
123 is not an armstrong number

CreatedOct 15, 2014

UpdatedDec 30, 2019

Views4,157
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