Smallest Even Multiple
The task is to find the smallest positive integer that is a multiple of both 2 and the given number n
.
A key observation is that for an integer to be divisible by 2, it must be an even number. Therefore, if n
is an even number, it’s already the smallest possible number that is a multiple of both 2 and n
. However, if n
is an odd number, the smallest number that is a multiple of both 2 and n
would be 2*n
.
We can write this logic in Python as follows:


In this function, if n
is an even number (i.e., n % 2 == 0
), it simply returns n
. If n
is odd, it returns 2 * n
. This ensures that the smallest even multiple of n
is returned.
This problem contradicts itself.
The problem asks us to find the smallest positive integer that is a multiple of both 2 and n.
Example 2 notes that a number is a multiple of itself, and then gives the answer for 6 as 6.
However, if a number is a multiple of itself, then the correct answer is 2, as 2 is a multiple of itself and 6.
When the problem statement says that the result needs to be a multiple of both 2 and n
, it implies that the result needs to be a multiple of n
, but it also has to be an even number.
Let’s take the example where n
is 6.
6 is an even number and a multiple of itself. So, it satisfies the conditions mentioned in the problem statement  it’s an even number and a multiple of n
.
If you consider 2 as a result, it is indeed a multiple of 2 but it’s not a multiple of n
(6 in this case).
The key point is that the number we’re looking for has to be a multiple of n
(and hence could be n
itself if n
is even), but it also has to be an even number.
This is why, if n
is already an even number, n
is returned. If n
is an odd number, 2*n
is returned to ensure the result is even.


The C++ function smallestEvenMultiple
calculates the smallest multiple of n
that is even. It takes an integer n
as an argument and returns its smallest even multiple.
Here’s how the function works:
It calculates
n % 2 + 1
. Ifn
is even,n % 2
is 0, and this expression results in 1. Ifn
is odd,n % 2
is 1, and this expression results in 2.It then multiplies
n
by the result of the above calculation. Ifn
was even, this results inn * 1
, which isn
itself. Ifn
was odd, this results inn * 2
, which is the smallest even multiple ofn
.
The function uses simple arithmetic operations and conditional logic to find the smallest even multiple of n
, resulting in a time complexity of O(1) and a space complexity of O(1).


The C++ function smallestEvenMultiple
calculates the smallest even multiple of n
. It does this by calculating the least common multiple (LCM) of 2
and n
using the lcm
function.
Here’s how the function works:
 The function
lcm(2, n)
calculates the least common multiple of2
andn
. The least common multiple of two numbers is the smallest number that is a multiple of both numbers. Since2
is the smallest even number, the LCM of2
and any numbern
will be the smallest even multiple ofn
.
This function uses the lcm
function to find the smallest even multiple of n
, resulting in a time complexity of O(log n), where n is the input number. The space complexity is O(1).
The lcm
function is not shown here, and it should be implemented separately. The function typically utilizes the mathematical relationship lcm(a, b) = a*b / gcd(a, b)
, where gcd(a, b)
is the greatest common divisor of a
and b
.