## Introduction to Armstrong Number in C++

A number that is equal to the sum of the cube of its digit is an Armstrong Number. A number is called as an Armstrong number if the sum of cube of its all digit is equal to that number. In this article, we are going to discussed how to check the Armstrong number using the C++ programming language. Some of the Armstrong numbers are – 0, 1, 153, 407. Let’s check it using mathematical computation.

0 = 0 * 0 * 0 = 0

1 = 1 * 1 * 1= 1

153 = (1 * 1 * 1) + (5 * 5 * 5) + (3 * 3 * 3) = 1 + 125 + 27 = 153

407 = (4 * 4 * 4) + (0 * 0 * 0) + (7 * 7 * 7) = 64 + 0 + 343 = 407

**Algorithm to Check Armstrong Number**

The algorithm to check armstrong number in C++ are given below:

**Step 1:** Enter Number

**Step 2:** Find the cube of each digit of entered number

**Step 3:** Add the cube of all the digits

**Step 4:** If the output of step 3 is equal to the entered number i.e. Step 1. Then the print entered number is Armstrong number.

**Step 5:** If the output of step 3 is equal to the entered number i.e. Step 1. Then print entered number is not an Armstrong number.

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### Examples of Armstrong Number

In this section, we are going to discussed how to check Armstrong’s number using various methods.

#### Example #1 – Program to check Armstrong number using a while loop.

**Code:**

`#include <iostream>`

using namespace std;

int main()

{

int num, r, sum=0, temp_num;

cout << "Enter number to check Armstrong number ";

cin >> num;

temp_num = num;

while(num > 0)

{

r = num % 10;

sum = sum + (r * r * r);

num = num / 10;

}

if(temp_num == sum)

cout << "Entered number is Armstrong Number." << endl;

else

cout << "Entered number is not Armstrong Number." << endl;

return 0;

}

**Output:**

Here we have written a program to check Armstrong number using a while loop, first it asks a user to enter a value. Then the entered number is copied into temp_num. Here temp_num is used to compare the result with the original. while condition checks whether the number is greater than 0 or not. If the number is greater than 0, it executes the statements following a while. The last digit is separated from num by performing num%10. Then the digit is cubed and stored the sum. Then the last digit is discarded using num/10. The process is performed for all digit in the number. Then temp_num and num are compared, if both are equal it will print Entered number is Armstrong Number. If both are not equal it will print Entered number is not Armstrong Number.

#### Example #2 – Program to Check Armstrong Number using a do-while loop.

**Code:**

`#include <iostream>`

using namespace std;

int main()

{

int num, r, sum=0, temp_num;

cout << "Enter number to check Armstrong number ";

cin >> num;

temp_num = num;

do

{

r = num % 10;

sum = sum + (r * r * r);

num = num / 10;

} while(num > 0);

if(temp_num == sum)

cout << "Entered number is Armstrong Number." << endl;

else

cout << "Entered number is not Armstrong Number." << endl;

return 0;

}

**Output:**

Here we have written a program to check Armstrong’s number using the do-while loop. The working is the same as we have discussed in example 1. The only difference is in 1st example if first checks the condition i.e num > 0. and here in this example the same condition is tested at the end of the loop.

#### Example #3 – Program to print Armstrong Number using for loop.

**Code:**

`#include <iostream>`

using namespace std;

int main()

{

int lower_limit, upper_limit, i, r, sum, temp_num;

cout << "Enter lower limit ";

cin >> lower_limit;

cout << "Enter uppee limit ";

cin >> upper_limit;

cout << "List of Armstrong numbers between " << lower_limit << " and " << upper_limit << endl;

for(i = lower_limit; i <= upper_limit; i++)

{

sum = 0;

temp_num = i;

for(; temp_num >0; temp_num /= 10)

{

r = temp_num % 10;

sum = sum + (r * r * r);

}

if(sum == i)

cout << i << endl;

}

return 0;

}

**Output:**

Here we have written a program to print the Armstrong number between two numbers entered by the users. The lower limit takes the minimum number and the upper limit takes the maximum number. If the upper limit number is small then the lower limit then it throws an error. The upper limit number should be greater than the lower limit. Each number between the interval is stored in temp_num. Then each digit of the number is retrieved in variable r and then finds the cube. The result of the cube is then added to the result of the last digit. Likewise, each digit is traversed, when traversing is done, the sum is compared with the original number i.e. i. If they are equal then it prints the number.

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