6Perfect
- Proper divisors
- 1, 2, 3
- Sum
- 6
Check Perfect Number in JavaScript
What you’ll learn
- What a perfect number is, and how it sits beside deficient and abundant numbers.
- How to sum proper divisors in JavaScript with a simple loop and the
%operator. - A live preview that shows divisors and the verdict before you run code snippets.
- Two complete programs: test one value and list perfect numbers in a range.
Overview
List every proper divisor of n (all positive divisors less than n) and add them. If the total equals n, the number is perfect. The smallest example is 6, because 1 + 2 + 3 = 6.
Two programs
One checks a single value; another scans a small range and prints all perfect hits.
Live preview
Type a number and instantly see divisor list, sum, and perfect/abundant/deficient classification.
Interview-ready
Clear loop logic and edge-case handling for n = 1.
Prerequisites
Comfort with loops, modulo arithmetic, and simple helper functions.
- Basic JavaScript syntax, functions, and
console.log. - Understanding that
n % i === 0meansidividesnevenly.
What is a perfect number?
A perfect number is a positive integer n whose proper divisors add up exactly to n.
With s(n) as proper-divisor sum:
Deficients(n) < n
Perfects(n) = n
Abundants(n) > n
Same idea, two formulas
If σ(n) is sum of all divisors (including n), then s(n) = σ(n) - n. So saying s(n) = n is the same as saying σ(n) = 2n.
Numbers to try in your head
28Perfect
- Proper divisors
- 1, 2, 4, 7, 14
- Sum
- 28
10Deficient
- Proper divisors
- 1, 2, 5
- Sum
- 8
12Abundant
- Proper divisors
- 1, 2, 3, 4, 6
- Sum
- 16
Live preview
Walks from 1 to n/2, sums proper divisors, and classifies the result.
Algorithm
Goal: decide whether positive integer n is perfect via proper-divisor sum.
Initialize sum = 0
Running total of proper divisors.
Loop i = 1 .. floor(n / 2)
If n % i === 0, add i to sum.
Compare sum and n
sum === n means perfect.
📜 Pseudocode
Pseudocode
function properDivisorSum(n):
sum = 0
for i from 1 to floor(n / 2):
if n mod i == 0:
sum += i
return sum
function isPerfect(n):
if n < 2: return false
return properDivisorSum(n) == n1
Check one number
Classic helper-based solution with 28 as sample.
JavaScript
function isPerfectNumber(number) {
let sum = 0;
for (let i = 1; i <= Math.floor(number / 2); i++) {
if (number % i === 0) {
sum += i;
}
}
return sum === number;
}
const number = 28;
if (isPerfectNumber(number)) {
console.log(number + " is a perfect number.");
} else {
console.log(number + " is not a perfect number.");
}2
Perfect numbers in a range
Reuse same helper over a small interval, e.g. 1 to 50.
JavaScript
function isPerfectNumber(num) {
let sum = 0;
for (let i = 1; i <= Math.floor(num / 2); i++) {
if (num % i === 0) {
sum += i;
}
}
return sum === num;
}
const values = [];
for (let i = 1; i <= 50; i++) {
if (isPerfectNumber(i)) {
values.push(String(i));
}
}
console.log("Perfect Numbers in the range 1 to 50:");
console.log(values.join(" "));Notes
Faster scan. You can iterate to Math.floor(Math.sqrt(n)) and add divisor pairs to reduce runtime.
Large numbers. JavaScript uses floating-point numbers; very large integers may lose exact precision.
Rarity. Perfect numbers are sparse, so scanning huge ranges is expensive.
❓ FAQ
A positive integer is perfect when the sum of all its proper divisors (excluding itself) equals the number. Example: 6 = 1 + 2 + 3.
No. Its proper-divisor sum is treated as 0, so it does not equal 1.
With s(n) as proper-divisor sum: s(n) > n is abundant, s(n) < n is deficient, and s(n) = n is perfect.
No proper divisor can exceed n/2, so the check is complete while staying simple for beginners.
Yes. You can loop to sqrt(n) and add divisor pairs, reducing time for one check from O(n) to O(sqrt(n)).
No, they are rare. That is why beginner examples usually test 28 or scan only a small range.
🔄 Input / output examples
| Value tested | Typical output |
|---|---|
| 28 | 28 is a perfect number. |
| 10 | 10 is not a perfect number. |
| 6 | 6 is a perfect number. |
| 12 | 12 is not a perfect number. |
Edge cases
n = 1Not perfect
Proper-divisor sum is 0, so it does not match 1.
Prime numbers
Always deficient
Only proper divisor is 1, so sum is less than the number.
Do not add n
Proper divisors only
Including n itself breaks the definition.
⏱️ Time and space complexity
| Task | Time | Extra space |
|---|---|---|
One check (loop to n/2) | O(n) | O(1) |
| One check (sqrt pairing) | O(sqrt(n)) | O(1) |
Scan 1..U naively | O(U^2) | O(1) |
Summary
- Definition: perfect when proper-divisor sum equals the number.
- Pattern: loop with
%, accumulatesum, compare withn. - Care: handle
n = 1, large ranges, and precision for huge values.
Did you know?
The first four perfect numbers are 6, 28, 496, and 8128.
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