Sum of Factors of a Number

Given a number n, find the sum of all the factors.

Examples :

Input: n = 30
Output: 72
Explanation: Divisors sum 1 + 2 + 3 + 5 + 6 + 10 + 15 + 30 = 72

Input: n = 15
Output: 24
Explanation: Divisors sum 1 + 3 + 5 + 15 = 24

Iterative Method - O(√n) Time and O(1) Space

Traverse from 2 to √n because factors always occur in pairs (i, n / i), so we can find all divisors till √n. Initialize the res with 1 (already including factor 1). For every divisor found, add both i and (n / i), handling the perfect square case where both are equal. Finally, add n to get the total sum.

For n = 15:

• Initialize res = 1 (including factor 1)
• √15 ≈ 3, so loop runs from i = 2 to 3
• i = 2, 15 is not divisible by 2, so skip
• i = 3, 15 is divisible by 3, so factors are 3 and 5, res becomes 1 + 3 + 5 = 9
• Add n itself, res becomes 9 + 15 = 24

Final Answer = 24

72

Keep Dividing n with Prime Factors

Use the prime factorization of n and apply the geometric progression formula to compute the sum of divisors efficiently. Each prime factor contributes a series (1 + p + p² + ... + p^a), and multiplying all such series gives the final sum.

If, n = p₁^a1 x p₂^a2 x ... x p_k^ak

Then, Sum of factors = (1 + p₁ + p₁² + ... + p₁^a1) x (1 + p₂ + p₂² + ... + p₂^a2) x ... x (1 + p_k + p_k² + ... + p_k^ak)

We can notice that individual terms of the above formula are Geometric Progressions (GP). So, we can rewrite it as:

Consider the number n = 18

18 = 2¹ × 3²
Factors = 1, 2, 3, 6, 9, 18

Consider each prime Factor and add all its powers divisible by n
Sum = (2⁰ + 2¹) × (3⁰ + 3¹ + 3²)
= (1 + 2) × (1 + 3 + 9)
= (1 + p₁) × (1 + p₂ + p₂²)

So the problem reduces to finding the prime factors of n along with their powers.

72