Question

How to check whether a number is a perfect number in Kotlin

Answer 1

// If the sum of all factors of a number is equal to the number, then the number is perfect
  
// Factors of a number are numbers that divide the number evenly
     
// 6
// factors = 1, 2, 3
// 1 + 2 + 3 = 6

fun main() {
    val num = 6
    var sumOfFactors = 0

    for (i in 1 until num) {
        if (num % i == 0) {
            sumOfFactors += i
        }
    }

    if (sumOfFactors == num) {
        println("Perfect Number")
    } else {
        println("Not a Perfect Number")
    }
}


 
 
/*
run:
   
Perfect Number
   
*/

 

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Answer 2

// If the sum of all factors of a number is equal to the number, then the number is perfect
  
// Factors of a number are numbers that divide the number evenly
     
// 6
// factors = 1, 2, 3
// 1 + 2 + 3 = 6

fun isPerfect(n: Int): Boolean {
    if (n < 2) return false

    var sum = 1
    val limit = kotlin.math.sqrt(n.toDouble()).toInt()

    for (i in 2..limit) {
        if (n % i == 0) {
            sum += i
            val pair = n / i
            if (pair != i) sum += pair
        }
    }

    return sum == n
}

fun main() {
    println("Perfect numbers up to 10,000:")
    for (i in 1..10000) {
        if (isPerfect(i)) println(i)
    }
}


/*
run:

Perfect numbers up to 10,000:
6
28
496
8128

*/

 

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