The "Smallest Even Multiple" interview question is a basic mathematical challenge that asks you to find the smallest positive integer that is a multiple of both 2 and a given integer n. Essentially, you are looking for the Least Common Multiple (LCM) of 2 and n. This "Smallest Even Multiple coding problem" is often used as a warm-up exercise to test a candidate's ability to handle simple conditional logic and mathematical properties.
Companies like Amazon and Google use this question in early-stage screenings or for entry-level positions. It tests if a candidate understands the concept of parity (even vs. odd) and basic multiples. While simple, it requires a moment of logical deduction: if a number is already even, it is already a multiple of 2, so the smallest even multiple is the number itself. If it's odd, you must multiply it by 2.
This problem follows the "Math and Number Theory interview pattern". The logic is a simple conditional:
n % 2 == 0 (n is even), the result is n.n % 2 != 0 (n is odd), the result is n * 2.
This is a constant time O(1) solution that requires no loops or complex data structures.Example 1: Let n = 6.
Since 6 is even, its multiples are 6, 12, 18... The smallest one is 6. Since 6 is also a multiple of 2, the answer is 6.
Example 2: Let n = 5.
Since 5 is odd, its multiples are 5, 10, 15... The smallest multiple of 5 that is also even is 10.
So the result for 5 is 10.
The most common mistake is overthinking the problem and implementing a loop to find the multiple. While a loop starting from n and checking i % 2 == 0 would work, it is less efficient than the direct conditional. Another mistake is forgetting that the result must be a multiple of both numbers, not just the smallest even number greater than n.
Always look for the simplest mathematical relationship before writing code. Many "EASY" level "Smallest Even Multiple interview question" scenarios can be solved with a single if statement or a ternary operator.