Sum Multiples - Practice Coding Problems

Sum Multiples - Problem

Math Easy

Given a positive integer n, find the sum of all integers in the range [1, n] inclusive that are divisible by 3, 5, or 7.

Return an integer denoting the sum of all numbers in the given range satisfying the constraint.

Input & Output

Example 1 — Basic Case

$ Input: n = 10

› Output: 40

💡 Note: Numbers divisible by 3, 5, or 7 in range [1,10]: 3, 5, 6, 7, 9, 10. Sum = 3+5+6+7+9+10 = 40

Example 2 — Smaller Range

$ Input: n = 7

› Output: 25

💡 Note: Numbers divisible by 3, 5, or 7 in range [1,7]: 3, 5, 6, 7. Sum = 3+5+6+7 = 21

Example 3 — Edge Case

$ Input: n = 3

› Output: 3

💡 Note: Only number 3 is divisible by 3, 5, or 7 in range [1,3]. Sum = 3

Constraints

1 ≤ n ≤ 10⁶

Visualization

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Understanding the Visualization

Input

Given positive integer n

Process

Find all numbers divisible by 3, 5, or 7

Output

Return sum of qualifying numbers

Key Takeaway

🎯 Key Insight: Use inclusion-exclusion principle to handle overlapping multiples efficiently

Asked in

G Google 25 a Amazon 18

The key insight is to use the inclusion-exclusion principle to avoid double counting numbers divisible by multiple factors. Best approach uses mathematical formulas with arithmetic series: Time: O(1), Space: O(1)

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force	O(n)	O(1)	Check each number from 1 to n and add if divisible by 3, 5, or 7
Mathematical Formula (Inclusion-Exclusion)	O(1)	O(1)	Use arithmetic series formula with inclusion-exclusion principle

Brute Force — Algorithm Steps

Loop through numbers 1 to n
Check if number % 3 == 0 OR % 5 == 0 OR % 7 == 0
Add qualifying numbers to sum

Visualization

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Step-by-Step Walkthrough

Start Loop

Check numbers 1 through n

Test Divisibility

For each number, test if divisible by 3, 5, or 7

Sum Valid Numbers

Add qualifying numbers to running total

Code -

solution.c — C

#include <stdio.h>

int solution(int n) {
    int total = 0;
    for (int i = 1; i <= n; i++) {
        if (i % 3 == 0 || i % 5 == 0 || i % 7 == 0) {
            total += i;
        }
    }
    return total;
}

int main() {
    int n;
    scanf("%d", &n);
    int result = solution(n);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single loop through n numbers

✓ Linear Growth

Space Complexity

O(1)

Only using constant extra variables

✓ Linear Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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