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Sum of Divisors Calculator

Mathematics - Sum of Divisors

Sum of Divisors Calculator — Calculate σ(n) Using Prime Factorization

Calculate the sum of all divisors of any positive integer with step-by-step explanation. Learn the σ(n) formula, understand proper divisors, and explore perfect, abundant, and deficient numbers. For related tools, see our divisor count calculator.

Table of Contents

  1. 1What Is the Sum of Divisors?
  2. 2The Divisor Sum Formula σ(n)
  3. 3How to Calculate σ(n)
  4. 4Step-by-Step Examples
  5. 5More Divisor Sum Examples
  6. 6Sum of Proper Divisors s(n)
  7. 7Perfect, Abundant, and Deficient Numbers
  8. 8The Sigma Function σₖ(n)
  9. 9Common Mistakes to Avoid
  10. 10Real-World Applications
  11. 11Practice Problems with Solutions
  12. 12Tips for Mastering Divisor Sums
  13. 13Final Thoughts
  14. 14FAQ: Common Questions

Calculate the Sum of Divisors

Enter a positive integer to find the sum of all its divisors. Click Calculate for step-by-step working.

Divisor sum will appear here.
Note: The divisor function σ(n) sums all positive divisors. Using prime factorization n = p₁ᵃ¹ × p₂ᵃ² × ..., the formula is: σ(n) = Π[(pᵢ^(aᵢ+1) - 1) / (pᵢ - 1)]. The sum of proper divisors is s(n) = σ(n) - n.

Sum of Divisors – Complete Explanation

The sum of divisors function σ(n) adds up all positive integers that divide evenly into n. The aliquot sum s(n) = σ(n) - n sums only proper divisors. These functions classify numbers as perfect, abundant, or deficient.

Divisor Sum Formula

If n = p₁ᵃ¹ × p₂ᵃ² × ... × pₖᵃᵏ

σ(n) = Π[(pᵢ^(aᵢ+1) - 1) / (pᵢ - 1)]

Example: 12 = 2² × 3 → σ(12) = [(2³-1)/(2-1)] × [(3²-1)/(3-1)] = 7 × 4 = 28.

1. What Is the Sum of Divisors?

σ(n) answers: "What is the total of all numbers that divide evenly into n?"

Key facts:

  • σ(1) = 1.
  • σ(p) = p + 1 for primes.
  • σ(n) = 2n for perfect numbers.
  • σ is multiplicative for coprime factors.

2. The Divisor Sum Formula Explained

For 12 = 2² × 3:

Sum of powers of 2: 1 + 2 + 4 = 7 = (2³-1)/(2-1)

Sum of powers of 3: 1 + 3 = 4 = (3²-1)/(3-1)

σ(12) = 7 × 4 = 28

Each divisor is formed by picking one term from each geometric series.

3. How to Calculate σ(n)

  1. Prime factorize n.
  2. For each prime power pᵉ: Calculate (pᵉ⁺¹ - 1)/(p - 1).
  3. Multiply all results — the product is σ(n).
  4. For proper divisors: s(n) = σ(n) - n.

4. Step-by-Step Examples

Example 1: σ(28) — Perfect Number

28 = 2² × 7
For 2²: (2³-1)/(2-1) = 7/1 = 7
For 7: (7²-1)/(7-1) = 48/6 = 8
σ(28) = 7 × 8 = 56
s(28) = 56 - 28 = 28 → PERFECT! ✨

Verification: 1+2+4+7+14+28 = 56 ✓

Example 2: σ(36)

36 = 2² × 3²
For 2²: 7
For 3²: (3³-1)/(3-1) = 26/2 = 13
σ(36) = 7 × 13 = 91
s(36) = 91 - 36 = 55 → ABUNDANT

Verification: 1+2+3+4+6+9+12+18+36 = 91 ✓

5. More Divisor Sum Examples

σ(6) = 12 — perfect

σ(12) = 28 — abundant

σ(10) = 18 — deficient

σ(17) = 18 — prime (17+1)

6. Sum of Proper Divisors s(n)

Aliquot Sum

s(n) = σ(n) - n

7. Perfect, Abundant, and Deficient Numbers

Perfect: s(n) = n → 6, 28, 496, 8128

Abundant: s(n) > n → 12, 18, 20, 24, 30, 36

Deficient: s(n) < n → most numbers, all primes

8. The Sigma Function σₖ(n)

σ₀(n) = d(n): Divisor count.

σ₁(n) = σ(n): Divisor sum.

σₖ(n): Sum of k-th powers of divisors.

9. Common Mistakes to Avoid

  • Confusing σ(n) with s(n).
  • Wrong formula for prime powers: σ(pᵉ) = (pᵉ⁺¹-1)/(p-1), not pᵉ+1 for e > 1.
  • Thinking σ(n) = n+1 for composites.
  • Forgetting σ(1) = 1.

10. Real-World Applications

  • Perfect number research — odd perfect numbers unsolved.
  • Cryptography — divisor analysis.
  • Aliquot sequences — iterating s(n).

11. Practice Problems with Solutions

Problem 1: σ(20) = ? → 20 = 2²×5 → 7×6 = 42

Problem 2: Is 18 perfect/abundant/deficient? → σ(18)=39, s(18)=21 → ABUNDANT

Problem 3: σ(100) = ? → 100 = 2²×5² → 7×31 = 217

Problem 4: σ(n) = 2n means? → Perfect number

12. Tips for Mastering Divisor Sums

  • Memorize the prime power formula.
  • Multiply σ of each prime power.
  • s(n) = σ(n) - n.
  • Perfect: σ(n) = 2n.

13. Final Thoughts

The sum of divisors function connects prime factorization to the classification of all positive integers. Use this calculator to verify your work, but practice applying the formula until it becomes automatic.

14. Frequently Asked Questions

σ(n) is the sum of all positive divisors of n, including 1 and n itself. For example, σ(12) = 1+2+3+4+6+12 = 28. It's a multiplicative arithmetic function central to number theory.

If n = p₁ᵃ¹ × p₂ᵃ² × ..., then σ(n) = Π[(pᵢ^(aᵢ+1) - 1)/(pᵢ - 1)]. For each prime power pᵉ, compute the geometric series sum (pᵉ⁺¹-1)/(p-1), then multiply all results.

σ(n) sums all divisors including n itself. s(n) = σ(n) - n is the aliquot sum — it sums only proper divisors (excluding n). For 12: σ(12) = 28, s(12) = 16.

When σ(n) = 2n, the number is a perfect number. This is equivalent to s(n) = n. Examples include 6 (σ(6)=12), 28 (σ(28)=56), and 496 (σ(496)=992).

For a prime p, the only divisors are 1 and p. Therefore, σ(p) = p + 1. The formula confirms this: (p²-1)/(p-1) = (p+1)(p-1)/(p-1) = p+1.

Using s(n) = σ(n) - n: If s(n) = n → perfect; if s(n) > n → abundant; if s(n) < n → deficient. The abundancy index σ(n)/n also classifies: = 2 perfect, > 2 abundant, < 2 deficient.

σ(n) sums the divisors; d(n) counts them. They are part of the same family: σ₀(n) = d(n) and σ₁(n) = σ(n). For 12: d(12) = 6 (six divisors), σ(12) = 28 (their sum).

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Conclusion

The sum of divisors function σ(n) connects prime factorization to number classification. Use this calculator to verify your work, but practice applying the formula until it becomes automatic.

Readers who wish to continue practicing and reinforce what they've learned can explore the companion learning resource for guided exercises.

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