## Q. Write a program to check whether the given number N is Armstrong number or not.

Armstrong Number :- A Armstrong Number is a number that is equal to sum of cubes of each digits.
For Example :- 153
So as we can see that the sum of cubes of each digit (1^3=1, 5^3=125, 3^3=27 => 1+125+27=153) is equal to it's number. So 153 is an Armstrong number.

## Armstrong Number Algorithm

```START
step 2 : set sum=0 and duplicate=number
step 3 : reminder=number%10
step 4 : sum=sum+(reminder*reminder*reminder)
step 5 : number=number/10
step 6 : repeat steps 4 to 6 until number > 0
step 7 : if sum = duplicate
step 8 : display number is armstrong
step 9 : else
step 10 : display number is not armstrong
STOP
```

# Armstrong Number Program

• C
• C++
• Java
• Python
• C#
• PHP
• ```#include <stdio.h>
int main() {
int num=153, new_num, rem, res = 0;
new_num = num;
while (new_num != 0) {
rem = new_num % 10;
res += rem * rem * rem;
new_num /= 10;
}
if (res == num)
printf("The number %d is an Armstrong number.", num);
else
printf("The number %d is not an Armstrong number.", num);
return 0;
}
```
```#include <iostream>
using namespace std;
int main()
{
int N=153, num, rem, sum = 0;
num = N;
while(num != 0)
{
rem = num % 10;
sum += rem * rem * rem;
num /= 10;
}
if(sum == N)
cout <<"The number "<< N <<" is an Armstrong number.";
else
cout <<"The number "<< N <<" is not an Armstrong number.";
return 0;
}
```
```public class LFC {
public static void main(String[] args) {
int num = 153, new_num, rem, res = 0;
new_num = num;
while (new_num != 0)
{
rem = new_num % 10;
res += Math.pow(rem, 3);
new_num /= 10;
}
if(res == num)
System.out.println("The number "+num+" is an Armstrong number.");
else
System.out.println("The number "+num+" is not an Armstrong number.");
}
}
```
```num = 153
# initialize sum
sum = 0
# find the sum of the cube of each digit
temp = num
while temp > 0:
digit = temp % 10
sum += digit ** 3
temp //= 10
# display the result
if num == sum:
print("The number {0} is Armstrong number".format(num))
else:
print("The number {0} is not Armstrong number".format(num))
```
```using System;
public class LFC
{
public static void Main(string[] args)
{
int  n,r,sum=0,temp;
n= 153;
temp=n;
while(n>0)
{
r=n%10;
sum=sum+(r*r*r);
n=n/10;
}
if(temp==sum)
Console.Write("The number "+temp+" is Armstrong Number.");
else
Console.Write("The number "+temp+" is Not Armstrong Number.");
}
}
```
```\$num=153;
\$total=0;
\$x=\$num;
while(\$x!=0)
{
\$rem=\$x%10;
\$total=\$total+\$rem*\$rem*\$rem;
\$x=\$x/10;
}
if(\$num==\$total)
{
echo "The number \$num is Armstrong number";
}
else
{
echo "The number \$num is not an Armstrong number";
}
```
```#include <stdio.h>
int main() {
int num=153, new_num, rem, res = 0;
new_num = num;
while (new_num != 0) {
rem = new_num % 10;
res += rem * rem * rem;
new_num /= 10;
}
if (res == num)
printf("The number %d is an Armstrong number.", num);
else
printf("The number %d is not an Armstrong number.", num);
return 0;
}
```

#### Output

```The number 153 is an Armstrong number.
```

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