The sum of three consecutive even integers problem asks you to find three even numbers that follow one another in sequence and add up to a specific total. These problems appear frequently in algebra courses and standardized tests like the SAT and ACT.
Even integers are whole numbers divisible by 2: ..., −4, −2, 0, 2, 4, 6, .... Consecutive even integers increase by 2 each time. For example, 10, 12, 14 are three consecutive even integers. Their sum is 36.
To solve such problems, assign a variable to the middle number. This simplifies equations because the three numbers are symmetric around the center. If the middle number is x, then the three numbers are x−2, x, and x+2. Their sum is (x−2) + x + (x+2) = 3x. Solving 3x = total gives the middle value.
Let the smallest even integer be represented by n. Since even numbers differ by 2, the next two consecutive even integers are n+2 and n+4. The sum becomes:
n + (n+2) + (n+4) = 3n + 6If the total sum is known, set this expression equal to it. For example, if the sum is 60:
3n + 6 = 60Solve for n:
The three numbers are 18, 20, and 22. Check: 18 + 20 + 22 = 60. Correct.
Using the middle number instead reduces steps. Let x be the middle even integer. Then the three numbers are x−2, x, x+2. Their sum is 3x. If the sum is 72:
3x = 72 → x = 24Numbers: 22, 24, 26. Sum: 72. Valid.
These problems often ask for the largest number, the smallest, or the product of the three. They may include negative integers or require checking divisibility.
| Type | Given | Variable Setup | Example |
|---|---|---|---|
| Find the integers | Sum = 48 | n + (n+2) + (n+4) = 48 | Solution: 14, 16, 18 |
| Find the largest | Sum = 90 | n + (n+2) + (n+4) = 90 | Largest = 32 |
| Include negatives | Sum = −18 | n + (n+2) + (n+4) = −18 | Integers: −8, −6, −4 |
A common mistake is assuming the first number is 2n instead of n. This leads to incorrect expressions. Always define the variable.
This problem type appears in high school algebra curricula and on college entrance exams. It teaches pattern recognition, equation setup, and verification skills.
According to the National Council of Teachers of Mathematics (NCTM), understanding sequences and algebraic modeling is essential for building quantitative reasoning. Problems like this help students move from arithmetic to symbolic thinking.
In real-world contexts, such logic applies to budgeting, scheduling, and programming loops. For example, a developer might need to process every second element in a list of even-numbered IDs.
People often mix up consecutive even integers with consecutive odd integers. The difference is the spacing: even integers differ by 2, but so do odd integers. However, their algebraic forms differ because even numbers are divisible by 2, while odds are not.
For consecutive odd integers, the setup is similar: n, n+2, n+4-but n must be odd. If you assume n is even, you'll get invalid results. For example, if the sum is 27, trying n=8 gives 8+10+12=30. Wrong. Using n=7: 7+9+11=27. Correct.
A non-obvious distinction: the sum of three consecutive even integers is always divisible by 6. This is because the sum is 3n + 6, which factors as 3(n+2). Since n is even, n+2 is even, so (n+2) is divisible by 2. Thus, 3 × (even number) is divisible by 6.
For example, 10+12+14=36. 36 ÷ 6 = 6. True. But 11+13+15=39. 39 ÷ 6 = 6.5. Not divisible. Confirms the rule.
Problem: The sum of three consecutive even integers is 132. What are the integers?
Let the middle number be x. Then the three numbers are x−2, x, x+2.
(x−2) + x + (x+2) = 132Simplify: 3x = 132 → x = 44.
So the integers are 42, 44, and 46. Check: 42 + 44 + 46 = 132. Verified.
I encountered this during a tutoring session with a 9th grader who struggled with setting up variables. After showing the symmetry around the middle number, they solved four similar problems correctly within 15 minutes.
When solving the sum of three consecutive even integers problem, use the middle number as the variable. This reduces complexity and minimizes errors. Always verify the solution by adding the numbers.