# Kotlin Program to Check Armstrong Number

#### In this program, you’ll learn to check whether a given number is armstrong number or not. You’ll learn to do this by using a while loop in Kotlin.

A positive integer is called an Armstrong number of order `n` if

abcd... = a^{n}+ b^{n}+ c^{n}+ d^{n}+ ...

In case of an Armstrong number of 3 digits, the sum of cubes of each digits is equal to the number itself. For example:

153 = 1*1*1 + 5*5*5 + 3*3*3 // 153 is an Armstrong number.

## Example 1: Check Armstrong Number for 3 digit number

```
fun main(args: Array<String>) {
val number = 371
var originalNumber: Int
var remainder: Int
var result = 0
originalNumber = number
while (originalNumber != 0) {
remainder = originalNumber % 10
result += Math.pow(remainder.toDouble(), 3.0).toInt()
originalNumber /= 10
}
if (result == number)
println("$number is an Armstrong number.")
else
println("$number is not an Armstrong number.")
}
```

When you run the program, the output will be:

371 is an Armstrong number.

- First, given number (
`number`)’s value is stored in another integer variable,`originalNumber`. This is because, we need to compare the values of final number and original number at the end. - Then, a while loop is used to loop through
`originalNumber`until it is equal to 0.- On each iteration, the last digit of
`num`is stored in`remainder`. - Then,
`remainder`is powered by 3 (number of digits) using`Math.pow()`

function and added to`result`.

Here,`remainder`is converted to`Double`

because`pow`

only accepts`Double`

parameters, and its value is again converted back to`Int`

- Then, the last digit is removed from
`originalNumber`after division by 10.

- On each iteration, the last digit of
- Finally,
`result`and`number`are compared. If equal, it is an armstrong number. If not, it isn’t.

## Example 2: Check Armstrong number for n digits

```
fun main(args: Array) {
val number = 1634
var originalNumber: Int
var remainder: Int
var result = 0
var n = 0
originalNumber = number
while (originalNumber != 0) {
originalNumber /= 10
++n
}
originalNumber = number
while (originalNumber != 0) {
remainder = originalNumber % 10
result += Math.pow(remainder.toDouble(), n.toDouble()).toInt()
originalNumber /= 10
}
if (result == number)
println("$number is an Armstrong number.")
else
println("$number is not an Armstrong number.")
}
```

In this program, we’ve used two while loops. The first while loop is used to count the number of digits in the `number`.

Then, `originalNumber` is restored to the given `number`.

The second while loop then checks whether the number is armstrong or not.

# Python Example for Beginners

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There are two sides to machine learning:

**Practical Machine Learning:**This is about querying databases, cleaning data, writing scripts to transform data and gluing algorithm and libraries together and writing custom code to squeeze reliable answers from data to satisfy difficult and ill defined questions. It’s the mess of reality.**Theoretical Machine Learning**: This is about math and abstraction and idealized scenarios and limits and beauty and informing what is possible. It is a whole lot neater and cleaner and removed from the mess of reality.

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