POTD Solutions | 21 Oct’ 23 | Sum of all divisors from 1 to n

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Welcome to the daily solutions of our PROBLEM OF THE DAY (POTD). We will discuss the entire problem step-by-step and work towards developing an optimized solution. This will not only help you brush up on your concepts of mathematics but will also help you build up problem-solving skills.

POTD 21 October: Sum of all divisors from 1 to n

Given a positive integer N., The task is to find the value of Σ_{i from 1 to N} F(i) where function F(i) for the number i is defined as the sum of all divisors of i.

Example:

Input: 4
Output: 15
Explanation:
F(1) = 1
F(2) = 1 + 2 = 3
F(3) = 1 + 3 = 4
F(4) = 1 + 2 + 4 = 7
ans = F(1) + F(2) + F(3) + F(4)
= 1 + 3 + 4 + 7
= 15

Input: 5
Output: 21

Sum of all divisors from 1 to n using basic mathematical concept:

The basic idea is to calculate the sum of divisors for a given number N by iterating through numbers from 1 to N and accumulating the contributions of each divisor to the total sum. For each number ‘i’ in the range, it calculates how many times ‘i’ divides ‘N’ without a remainder and adds this contribution to the sum. The final sum represents the sum of divisors for the input N.

Step-by-step approach:

• A long long variable called “sum” is initialized to zero. This variable will be used to accumulate the sum of divisors.
• A for loop iterates from 1 to N. This loop is used to find the divisors of the input number N.
• Inside the loop, the program calculates the contribution of each divisor to the sum. For each value of ‘i’ in the loop, it calculates (N / i) * i. Here’s how this works:
  • N / i gives the number of times ‘i’ divides ‘N‘ without a remainder. It essentially represents how many times ‘i’ appears as a divisor in ‘N‘.
  • Multiplying (N / i) by ‘i’ essentially gives the total value contributed by the divisor ‘i’ to the sum of divisors.
• The result of (N / i) * i is added to the “sum” variable, effectively accumulating the sum of divisors.
• Once the loop finishes iterating from 1 to N, the method returns the final value of the “sum,” which represents the sum of divisors of the input number N.

Below is the implementation of the above approach:

C++

class Solution {
public:
    // Function to calculate the sum of divisors of a given
    // number N.
    long long sumOfDivisors(int N)
    {
        long long sum = 0;

        // iterating from 1 to N.
        for (int i = 1; i <= N; ++i)
            // calculating and accumulating the sum of
            // divisors.
            sum += (N / i) * i;

        // returning the sum of divisors.
        return sum;
    }
};

Java

public class Solution {
    // Function to calculate the sum of divisors of a given
    // number N.
    public long sumOfDivisors(int N)
    {
        long sum = 0;

        // Iterating from 1 to N.
        for (int i = 1; i <= N; i++) {
            // Calculating and accumulating the sum of
            // divisors.
            sum += (N / i) * i;
        }

        // Returning the sum of divisors.
        return sum;
    }
}

Python3

class Solution:
    def sumOfDivisors(self, N):
        sum = 0

        # Iterating from 1 to N.
        for i in range(1, N + 1):
            # Calculating and accumulating the sum of divisors.
            sum += (N // i) * i

        # Returning the sum of divisors.
        return sum

Time complexity: O(N) 
Auxiliary space: O(1)