Smallest Even Multiple - Practice Coding Problems

Smallest Even Multiple - Problem

Given a positive integer n, you need to find the smallest positive integer that is a multiple of both 2 and n.

In other words, find the Least Common Multiple (LCM) of 2 and n.

Example: If n = 5, the multiples of 5 are [5, 10, 15, 20, ...] and the multiples of 2 are [2, 4, 6, 8, 10, 12, ...]. The smallest number that appears in both sequences is 10.

This is a fundamental number theory problem that helps you understand the relationship between LCM and GCD.

Input & Output

example_1.py — Basic Even Number

$ Input: n = 6

› Output: 6

💡 Note: Since 6 is already even, it's divisible by both 2 and 6. The LCM of 2 and 6 is 6.

example_2.py — Basic Odd Number

$ Input: n = 5

› Output: 10

💡 Note: 5 is odd, so we need the first multiple of 5 that's also even. That would be 5×2 = 10.

example_3.py — Edge Case Minimum

$ Input: n = 1

› Output: 2

💡 Note: 1 is odd, so the smallest multiple of both 1 and 2 is 1×2 = 2.

Constraints

1 ≤ n ≤ 150
n is a positive integer
The result will always fit in a 32-bit integer

Visualization

Tap to expand

Understanding the Visualization

Even Numbers

If n is even, it already contains factor 2, so LCM(2,n) = n

Odd Numbers

If n is odd, gcd(2,n) = 1, so LCM(2,n) = 2×n

Pattern

This creates a simple conditional: return n if even, 2n if odd

Key Takeaway

🎯 Key Insight: The parity (even/odd nature) of n directly determines the LCM with 2, making this an O(1) solution.

Asked in

G Google 15 a Amazon 12 ⊞ Microsoft 8 f Meta 6

The optimal solution uses the mathematical insight that LCM(2,n) = n if n is even, or 2×n if n is odd. This gives us O(1) time complexity by simply checking the parity of n.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Check Every Multiple)	O(1) to O(2)	O(1)	Check multiples of n until we find one divisible by 2
Mathematical Optimization (Optimal)	O(1)	O(1)	Use LCM formula: LCM(2,n) = n if n is even, 2×n if n is odd

Brute Force (Check Every Multiple) — Algorithm Steps

Start with multiple = n
Check if multiple is divisible by 2
If yes, return multiple
If no, increment to next multiple of n
Repeat until found

Visualization

Tap to expand

Step-by-Step Walkthrough

Start with n

Check if n is divisible by 2

If not, try 2n

2n is always even, so it will be our answer

Code -

solution.c — C

#include <stdio.h>

int smallestEvenMultiple(int n) {
    int multiple = n;
    while (multiple % 2 != 0) {
        multiple += n;
    }
    return multiple;
}

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

Time & Space Complexity

Time Complexity

⏱️

O(1) to O(2)

At most 2 iterations: if n is even, we find answer immediately; if n is odd, we check 2n

✓ Linear Growth

Space Complexity

O(1)

Only using a constant amount of extra space for variables

✓ Linear Space

73.5K Views

Medium Frequency

~5 min Avg. Time

2.5K Likes

Ln 1, Col 1

Smart Actions

💡 Explanation

AI Ready

💡 Suggestion Tab to accept Esc to dismiss

// Output will appear here after running code

Code Editor Closed

Click the red button to reopen

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force (Check Every Multiple) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Select Compiler