Check Abundant Number in JavaScript

What You’ll Learn

Abundant numbers are a classic interview warm-up: proper divisors, sum comparisons, and optional O(√n) speedups. This tutorial covers the definition, two JavaScript approaches with Try it Yourself editors, a live preview, algorithm steps, worked examples, edge cases, and complexity.

📝 Problem & Approach

Given a positive integer n, decide whether it is abundant; optionally list every abundant value in a closed interval.

// n = 12
// Proper divisors: 1, 2, 3, 4, 6
// s(12) = 16 > 12  → abundant
//
// Classification:
//   abundant  if s(n) > n
//   perfect   if s(n) = n
//   deficient if s(n) < n

Inputs & Outputs

Minimal workflow

function properDivisorSum(n):
    if n < 1:
        return invalid
    sum ← 0
    for i from 1 to floor(n / 2):
        if n mod i = 0:
            sum ← sum + i
    return sum

function isAbundant(n):
    return properDivisorSum(n) > n

Method comparison

🔮 Live Preview

Enter a positive integer and see proper divisors, how s(n) is added, and the abundant / perfect / deficient verdict.

Press “Run check”.

Examples Gallery

Two complete JavaScript programs — fast single check and a range scan. Use Try it Yourself to open an interactive editor, or View Output for the sample console result.

/**
 * Returns true if num is abundant (proper divisor sum > num).
 * Fast O(√num) scan using divisor pairs.
 */
function isAbundant(num) {
  if (num <= 1) {
    return false;
  }
  let sum = 1;

  for (let i = 2; i * i <= num; i++) {
    if (num % i === 0) {
      sum += i;
      const pair = Math.floor(num / i);
      if (i !== pair) {
        sum += pair;
      }
    }
  }

  return sum > num;
}

const number = 12;

if (isAbundant(number)) {
  console.log(number + " is an abundant number.");
} else {
  console.log(number + " is not an abundant number.");
}

12 is an abundant number.

function isAbundant(num) {
  if (num <= 1) {
    return false;
  }
  let sum = 0;

  for (let i = 1; i <= Math.floor(num / 2); i++) {
    if (num % i === 0) {
      sum += i;
    }
  }

  return sum > num;
}

process.stdout.write("Abundant numbers between 1 and 50 are: ");

for (let i = 1; i <= 50; i++) {
  if (isAbundant(i)) {
    process.stdout.write(i + " ");
  }
}

process.stdout.write("\n");

Abundant numbers between 1 and 50 are: 12 18 20 24 30 36 40 42 48

🔎 Worked Walkthrough — n = 12

Trace the naive proper-divisor sum for the smallest abundant number.

Compare: 16 > 12 → abundant. (Using σ: divisors sum to 28, and 28 > 24 = 2×12.)

🔄 Sample Values

Known results for the single-number script.

🎯 Practice Problems

Try these variations — use the Try it editors as a starting point.

Notes

⏱️ Time and Space Complexity

Both scripts on this page use only a few locals — auxiliary space is constant aside from the runtime.

💡 Best Practices