Let a

𝑎

 be an even integer. Then we always have gcd(a,a+2

𝑎

,

𝑎

+

2

)= ?

Question 7Answer

a.

2

2


b.

1

1


c.

4

4


d.

5

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.

Explore the properties of the greatest common divisor (gcd) when dealing with even integers. Understand how gcd(a, a+2) is determined and why it holds true. Suitable for high school students studying number theory and gcd concepts.

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