Let the four consecutive even integers be represented as:

• $x$, $x+2$, $x+4$, $x+6$

We are told that the sum of these integers equals the least of them, which is $x$. So, we set up the equation:

$x + (x + 2) + (x + 4) + (x + 6) = x$

Simplify the left-hand side:

$x + x + 2 + x + 4 + x + 6 = x$

$4x + 12 = x$

Now, subtract $x$ from both sides to isolate terms involving $x$:

$4x - x + 12 = 0$

$3x + 12 = 0$

Next, subtract 12 from both sides:

$3x = -12$

Finally, divide by 3 to solve for $x$:

$x = -4$

So, the four consecutive even integers are:

• $x = -4$
• $x+2 = -2$
• $x+4 = 0$
• $x+6 = 2$

Therefore, the integers are $-4, -2, 0, 2$.

Verification:

The sum of $-4 + (-2) + 0 + 2 = -4$, which matches the least integer, $-4$.

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Would you like a more detailed explanation? Let me know if you have any questions!

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Here are 5 follow-up questions to further explore the topic:

1. How would the solution change if the integers were consecutive odd numbers?
2. Can the sum of any set of consecutive integers be equal to the least integer?
3. What happens if we have an equation with three consecutive integers instead of four?
4. How would this problem change if the integers were multiples of 3 instead of even numbers?
5. What is the general formula for finding consecutive even integers given a specific condition like this?

Tip: When solving for consecutive integers, always express them in terms of a variable and use algebraic methods to form an equation. This approach makes it easier to handle problems like these!