Smallest Even Multiple

easy math number theory

Problem

Given a positive integer n, return the smallest positive integer that is a multiple of both 2 and n. In other words, find the least common multiple of n and 2.

Inputn = 5

Output10

5 is odd, so the smallest number divisible by both 5 and 2 is 5 × 2 = 10.

def smallest_even_multiple(n):
    if n % 2 == 0:
        return n
    return n * 2

function smallestEvenMultiple(n) {
  if (n % 2 === 0) {
    return n;
  }
  return n * 2;
}

class Solution {
    public int smallestEvenMultiple(int n) {
        if (n % 2 == 0) {
            return n;
        }
        return n * 2;
    }
}

int smallestEvenMultiple(int n) {
    if (n % 2 == 0) {
        return n;
    }
    return n * 2;
}

Explanation

We want the smallest positive number that both 2 and n divide evenly. That is exactly the least common multiple of n and 2, written lcm(n, 2).

There are only two cases to consider, and they hinge on whether n is even or odd.

If n is already even, then n is itself a multiple of 2, and nothing is smaller than n among its own multiples. So the answer is just n.

If n is odd, then n shares no factor of 2 with the number 2. To become divisible by 2 we must multiply by 2, giving 2 × n. This is the smallest even multiple because any smaller multiple of n would have to be n itself, which is odd.

Example: n = 5 is odd, so the answer is 5 × 2 = 10. For n = 6, which is already even, the answer is simply 6.

The whole thing is decided by a single parity check, so it runs in constant time.

Time: O(1) Space: O(1)