Given that $a$ is an even integer, we want to find the greatest common divisor (gcd) of $a$ and $a+2$.

Let's break it down:

• Since $a$ is even, we can express $a$ as $a = 2k$, where $k$ is an integer.
• The second term is $a + 2 = 2k + 2$, which can be factored as $2(k + 1)$.

So, the problem is asking for $\gcd(2k, 2k + 2)$.

Now, let's simplify:

1. $a = 2k$
2. $a + 2 = 2k + 2 = 2(k + 1)$

The gcd of two numbers can be determined by factoring out the common factors:

• $a = 2k$ has a factor of 2.
• $a + 2 = 2(k + 1)$ also has a factor of 2.

So, we factor out the common 2:

$\gcd(2k, 2(k + 1)) = 2 \times \gcd(k, k + 1)$

Since $k$ and $k + 1$ are consecutive integers, they are coprime, meaning their gcd is 1.

Therefore:

$\gcd(2k, 2k + 2) = 2 \times 1 = 2$

So, the correct answer is:

Option a. $\boxed{2}$

Would you like more details on how the gcd was calculated, or do you have any other questions?

Here are five related questions for further exploration:

1. What is the gcd of two consecutive integers?
2. How does the parity of a number affect its gcd with another number?
3. If $a$ is an odd integer, what would the gcd of $a$ and $a+2$ be?
4. How can the Euclidean algorithm be used to find the gcd of two numbers?
5. What is the significance of the gcd in number theory?

Tip: When working with gcd problems, breaking down the numbers into their prime factorizations can simplify the calculation process.