Problem 1: Multiples of 3 or 5

Problem Statement

If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.

Find the sum of all the multiples of 3 or 5 below 1000.

Approach

This problem can be solved using two approaches:

Brute Force Approach

Iterate through all numbers from 1 to 999 and check if each number is divisible by 3 or 5. If so, add it to the sum.

int sum = 0;
for (int i = 1; i < 1000; i++) {
    if (i % 3 == 0 || i % 5 == 0) {
        sum += i;
    }
}

Mathematical Approach (More Efficient)

Use the arithmetic series formula to calculate the sum of multiples directly:

• Sum of multiples of 3 below 1000
• Plus sum of multiples of 5 below 1000
• Minus sum of multiples of 15 below 1000 (to avoid double counting)

The sum of multiples of n below limit = n × (1 + 2 + 3 + … + k) where k = floor((limit-1)/n)

Complexity

• Time Complexity: O(n) for brute force, O(1) for mathematical approach
• Space Complexity: O(1)

Solution

View the complete solution on GitHub.

Related Concepts

• Modular arithmetic
• Arithmetic series
• Set theory (inclusion-exclusion principle)