How to resolve the algorithm Abundant odd numbers step by step in the Scala programming language
Published on 12 May 2024 09:40 PM
How to resolve the algorithm Abundant odd numbers step by step in the Scala programming language
Table of Contents
Problem Statement
An Abundant number is a number n for which the sum of divisors σ(n) > 2n, or, equivalently, the sum of proper divisors (or aliquot sum) s(n) > n.
12 is abundant, it has the proper divisors 1,2,3,4 & 6 which sum to 16 ( > 12 or n); or alternately, has the sigma sum of 1,2,3,4,6 & 12 which sum to 28 ( > 24 or 2n).
Abundant numbers are common, though even abundant numbers seem to be much more common than odd abundant numbers. To make things more interesting, this task is specifically about finding odd abundant numbers.
Let's start with the solution:
Step by Step solution about How to resolve the algorithm Abundant odd numbers step by step in the Scala programming language
Source code in the scala programming language
import scala.collection.mutable.ListBuffer
object Abundant {
def divisors(n: Int): ListBuffer[Int] = {
val divs = new ListBuffer[Int]
divs.append(1)
val divs2 = new ListBuffer[Int]
var i = 2
while (i * i <= n) {
if (n % i == 0) {
val j = n / i
divs.append(i)
if (i != j) {
divs2.append(j)
}
}
i += 1
}
divs.appendAll(divs2.reverse)
divs
}
def abundantOdd(searchFrom: Int, countFrom: Int, countTo: Int, printOne: Boolean): Int = {
var count = countFrom
var n = searchFrom
while (count < countTo) {
val divs = divisors(n)
val tot = divs.sum
if (tot > n) {
count += 1
if (!printOne || !(count < countTo)) {
val s = divs.map(a => a.toString).mkString(" + ")
if (printOne) {
printf("%d < %s = %d\n", n, s, tot)
} else {
printf("%2d. %5d < %s = %d\n", count, n, s, tot)
}
}
}
n += 2
}
n
}
def main(args: Array[String]): Unit = {
val max = 25
printf("The first %d abundant odd numbers are:\n", max)
val n = abundantOdd(1, 0, max, printOne = false)
printf("\nThe one thousandth abundant odd number is:\n")
abundantOdd(n, 25, 1000, printOne = true)
printf("\nThe first abundant odd number above one billion is:\n")
abundantOdd((1e9 + 1).intValue(), 0, 1, printOne = true)
}
}
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