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#### TO CHECK WHETHER A GIVEN NUMBER IS AN ARMSTRONG NUMBER OR NOT IN C

Code

```#include<stdio.h>
int main()
{
int num;
int rem=0,sum=0,n=0;
printf("ENTER THE NUMBER : ");
scanf("%d",&num);
n=num;
while(n>0)
{
rem=n%10;
sum=sum+rem*rem*rem;
n=n/10;
}
if(sum==num)
{
printf("IT IS AN ARMSTRONG NUMBER \n");
}
else
{
printf("IT IS NOT AN ARMSTRONG NUMBER \n");
}
return 0;
}
```

Output

ENTER THE NUMBER : 153
IT IS AN ARMSTRONG NUMBER What we did - Our Approach

In this problem we have to check whether a given number is an armstrong number or not.An Armstrong number is an integer such that the sum of the cubes of its digits is equal to the number itself. For example, 371 is an Armstrong number since 3*3*3 + 7*7*7 + 1*1*1 = 371. To do so we declare a variable 'sum',which will store the result value(don't forget to initialize it with zero).Then,we update our 'sum' value by adding cube of remainder value(rem*rem*rem) to it in each iteration(until 'n' becomes 0).
After the last iteration,we compare the value of given number and our sum ,if both values are equal then given number is armstrong else it isn't.

See this code in :

C++

Java

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