Check Perfect Number in Java
What you’ll learn
- How to decide if a number is perfect using proper divisors.
- How to implement single-check and range-check programs in Java.
- How perfect, abundant, and deficient numbers differ.
What is a perfect number?
A positive integer is perfect when the sum of its proper divisors equals the number itself.
Example: for 6, proper divisors are 1, 2, 3 and 1 + 2 + 3 = 6, so 6 is perfect.
Prerequisites
Divisors, modulo checks, and simple loop-based accumulation.
- Divisors and modulo operation.
- Looping from
1ton/2.
The idea
Add all proper divisors of n. If the sum equals n, it is perfect.
Definition
A number is perfect if the sum of its proper divisors equals the number: s(n) = n.
Intuition examples
6 -> 1+2+3=6 (perfect), 12 -> 1+2+3+4+6=16 (abundant).
Live preview
Type a value and check whether it is perfect, abundant, or deficient.
Algorithm
Initialize sum
Start with sum = 0.
Scan divisors
Loop i from 1 to n / 2.
Add proper divisors
If n % i == 0, add i to sum.
Compare
If sum == n, number is perfect.
📜 Pseudocode
function isPerfect(n):
if n <= 1:
return false
sum = 0
for i from 1 to floor(n / 2):
if n mod i == 0:
sum = sum + i
return sum == nCheck one number
This example checks whether 28 is perfect.
public class Main {
static boolean isPerfectNumber(int number) {
if (number <= 1) return false;
int sum = 0;
for (int i = 1; i <= number / 2; i++) {
if (number % i == 0) {
sum += i;
}
}
return sum == number;
}
public static void main(String[] args) {
int number = 28;
if (isPerfectNumber(number)) {
System.out.println(number + " is a perfect number.");
} else {
System.out.println(number + " is not a perfect number.");
}
}
}Perfect numbers in a range
This prints all perfect numbers from 1 to 50.
public class Main {
static boolean isPerfectNumber(int number) {
if (number <= 1) return false;
int sum = 0;
for (int i = 1; i <= number / 2; i++) {
if (number % i == 0) {
sum += i;
}
}
return sum == number;
}
public static void main(String[] args) {
System.out.println("Perfect numbers between 1 and 50:");
for (int i = 1; i <= 50; i++) {
if (isPerfectNumber(i)) {
System.out.print(i + " ");
}
}
}
}Optimization note
You can iterate to sqrt(n) and add divisor pairs to reduce time.
❓ FAQ
🔄 Input / output examples
| n | Verdict |
|---|---|
28 | Perfect |
10 | Not perfect |
6 | Perfect |
Edge cases
1 is not perfect
Its proper-divisor sum is 0, not 1.
Large values
For larger ranges, use long to avoid sum overflow.
⏱️ Time and space complexity
| Task | Time | Extra space |
|---|---|---|
| Check one number | O(n) | O(1) |
| Check one number with sqrt optimization | O(√n) | O(1) |
Summary
- Compute proper-divisor sum and compare with
n. - This direct method is
O(n); sqrt-pairing improves it.
The first four perfect numbers are 6, 28, 496, and 8128.
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