6Perfect
- Divisors
- 1, 2, 3
- Sum
- 6
Check Perfect Number in Python
What you’ll learn
- What perfect, deficient, and abundant numbers mean.
- How to sum proper divisors in Python.
- How to check one number and scan a range.
Overview
A perfect number equals the sum of its proper divisors. For example, 6 is perfect because 1 + 2 + 3 = 6.
Two programs
One single-value check and one range scan.
Live preview
See divisors, sum, and classification immediately.
Interview-ready flow
Clear loop, edge case for 1, and complexity discussion.
Prerequisites
Python loops, modulo operator, and basic function syntax.
- Use
for i in range(...)loops comfortably. - Know that
n % i == 0means i divides n.
What is a perfect number?
A positive integer n is perfect when the sum of its proper divisors equals n exactly.
Using s(n) = sum of proper divisors:
Deficient s(n) < n
Perfect s(n) = n
Abundant s(n) > n
Equivalent formulas
If sigma(n) is the sum of all positive divisors, then s(n) = sigma(n) - n. So s(n) = n is equivalent to sigma(n) = 2n.
Checklist
Compute proper divisor sum and compare with n.
Quick examples
28Perfect
- Divisors
- 1, 2, 4, 7, 14
- Sum
- 28
10Deficient
- Divisors
- 1, 2, 5
- Sum
- 8
12Abundant
- Divisors
- 1, 2, 3, 4, 6
- Sum
- 16
Live preview
Enter a positive integer and inspect proper divisors, sum, and verdict.
Algorithm
Goal: check if proper divisor sum equals n.
Initialize sum
Start with sum = 0.
Scan divisors
For i from 1 to n // 2, if n % i == 0 then add i.
Compare
If sum == n, number is perfect.
📜 Pseudocode
Pseudocode
function proper_divisor_sum(n):
sum = 0
for i from 1 to floor(n / 2):
if n mod i == 0:
sum = sum + i
return sum
function is_perfect(n):
if n < 2:
return false
return proper_divisor_sum(n) == n1
Check one number
Test a fixed value (28) with a helper function.
python
def is_perfect_number(number: int) -> bool:
total = 0
for i in range(1, number // 2 + 1):
if number % i == 0:
total += i
return total == number
number = 28
if is_perfect_number(number):
print(f"{number} is a perfect number.")
else:
print(f"{number} is not a perfect number.")2
Perfect numbers in range 1 to 50
Reuse helper and print all perfect numbers in a small interval.
python
def is_perfect_number(num: int) -> bool:
total = 0
for i in range(1, num // 2 + 1):
if num % i == 0:
total += i
return total == num
print("Perfect Numbers in the range 1 to 50:")
for i in range(1, 51):
if is_perfect_number(i):
print(i, end=" ")Notes
Faster check. You can test divisors only up to sqrt(n) and add divisor pairs.
Rarity. Perfect numbers are rare, so broad brute-force scans are expensive.
❓ FAQ
A positive integer is perfect if the sum of its proper divisors equals the number itself.
No. Its proper divisor sum is treated as 0, not 1.
If proper divisor sum is less than n it is deficient, if greater than n it is abundant.
No proper divisor of n can be larger than n // 2.
Yes. It is mathematically equivalent to proper-divisor sum equals n.
No, they are very rare.
🔄 Input / output examples
Replace fixed values in code or read from input for interactive runs.
| Value | Output idea |
|---|---|
| 28 | Perfect number |
| 10 | Not perfect |
| 6 | Perfect number |
| 12 | Not perfect |
Edge cases
n = 1
Not perfect
Proper divisor sum is 0, not 1.
Prime numbers
Always deficient
Only proper divisor is 1, so sum is less than n.
Do not include n
Proper divisors only
Adding n itself breaks the definition.
⏱️ Time and space complexity
| Task | Time | Extra space |
|---|---|---|
| Single check with n // 2 scan | O(n) | O(1) |
| Single check with sqrt(n) pairing | O(sqrt(n)) | O(1) |
| Range scan 1..U (naive) | O(U^2) | O(1) |
Summary
- Definition: perfect means proper divisor sum equals n.
- Loop: check divisors from 1 to n // 2 with modulo.
- Edge case: 1 is not perfect.
Did you know?
The first four perfect numbers are 6, 28, 496, and 8128. Whether any odd perfect number exists is still unknown.
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