Kotlin program to check Armstrong number

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In this program, you will learn to check if a given number is an Armstrong number. You will learn how to do this by using a while loop in Kotlin.

A positive integer is called an Armstrong number if

abcd... = an　+　bn　+　cn　+　dn　+　...

For3An Armstrong number of n digits, the sum of the cubes of each digit equals the number itself. For example:

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

Example1：Check if a number is an Armstrong number of3digit

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　+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 the program is run, the output is:

371　is an Armstrong number

• First, the value of the given number (number) is stored in another integer variable originalNumber. This is because we need to compare the final number with the original number's value at the end.

• Then, use a while loop to traverse originalNumber until it equals 0.

• In each iteration, the last digit of num is stored in remainder.

• Then, add remainder by using the Math.pow() function3（number of digits）and add it to result.

• Then, divide by10Then, remove the last digit from originalNumber.
  

• Finally, compare result and number. If they are equal, then it is an Armstrong number. If not, it is not.

The equivalent Java code is as follows:Java program to check Armstrong number

Example2：Check if a number is an Armstrong number of 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　+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 used two while loops. The first while loop is used to calculate the number of digits in number.

Then, originalNumber is restored to the given number.

Then, the second while loop checks if the number is an Armstrong number.

Visit this page to learn howDisplay All Armstrong Numbers Between Two Time Intervals.

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