Let's solve the problem step by step.

(a) Proof: The sum of any three consecutive even integers is a multiple of 6.

Let the three consecutive even integers be:

• $x$ (the first integer),
• $x+2$ (the second integer),
• $x+4$ (the third integer).

The sum of these three integers is: $x + (x+2) + (x+4) = 3x + 6.$

Factoring out a 3: $3x + 6 = 3(x + 2).$

Since $x+2$ is an integer (because $x$ is an integer), $3(x+2)$ is a multiple of 3. Additionally, since $x$ is even, $x+2$ is also even, meaning $x+2$ is divisible by 2. Therefore: $3(x+2)$ is divisible by both 2 and 3, meaning it is a multiple of 6.

Thus, the sum of three consecutive even integers is always a multiple of 6.

(b) Values of three consecutive even integers if their sum is 6000.

Using the general formula for the sum of three consecutive even integers: $3x + 6 = 6000.$

Simplify: $3x = 6000 - 6,$ $3x = 5994.$

Divide by 3: $x = \frac{5994}{3} = 1998.$

Thus, the three consecutive even integers are: $x = 1998, \quad x+2 = 2000, \quad x+4 = 2002.$

Final Answer:

(a) The sum of any three consecutive even integers is a multiple of 6.

(b) The three consecutive even integers whose sum is 6000 are: $1998, 2000, \text{ and } 2002.$

Would you like me to explain any step further? Here are 5 related questions for exploration:

1. How can you generalize this proof for odd integers instead of even integers?
2. What if the integers were consecutive multiples of 3? Would the result still hold?
3. How would the formula change for 4 consecutive integers instead of 3?
4. Can you prove that the sum of any $n$ consecutive integers is divisible by $n$ if $n$ is odd?
5. What mathematical property ensures that multiples of 2 and 3 are multiples of 6?

Tip: Always check for common factors in arithmetic expressions when proving divisibility properties!