Kotlin Program to Check Armstrong Number

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

abcd... = an + bn + cn + dn + ...

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.
• Finally, result and number are compared. If equal, it is an armstrong number. If not, it isn't.

Here's the equivalent Java code: Java Program to Check Armstrong Number

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.

Visit this page to learn, how you can display all armstrong numbers between two intervals.