Consecutive Even Integers: Sum Equals Four Times the Smallest
Abigail wants to find three consecutive even integers whose sum is four times the smallest of those integers. She lets n represent the smallest integer, then writes this equation: n+(n+2)+(n+4)=4n. What are the three integers?
Already Got the Answer? Check It Here
Setup:n + (n+2) + (n+4) = 4n simplifies to 3n + 6 = 4n.
Answer:n = 6, so the integers are 6, 8, 10.
One-line check:6 + 8 + 10 = 24 and 4 × 6 = 24 ✓
If you have the answer, jump to the pitfalls, the pattern section and the four What-If problems below. That is where the page goes beyond the solution.
What You Will Learn
- Reading an equation someone else wrote. Abigail has already done the modeling. Your job is to understand why each piece of her equation is there, and that skill matters as much as writing your own.
- Why "consecutive even" means "add 2." The gap between consecutive terms depends on the type of integer, and getting that gap right is the heart of every problem in this family.
- Solving with the variable on both sides. Collecting
nterms on one side is a core technique, and here it produces a surprisingly tidy result. - Finishing the question that was asked. Solving for
nis not the same as answering "what are the three integers?" - Choosing a smarter variable. Centering the integers around the middle value can make the algebra almost disappear.
Solution: Method 1 — Build Everything from the Smallest Integer
This is Abigail's own route, and it is the standard way to handle consecutive-integer problems: pick one variable for the smallest number and express every other number in terms of it.
Step 1 — Understand what each piece of the equation means
Abigail lets n be the smallest integer. Even integers are two apart (2, 4, 6, 8, …), so the next even integer after n is n + 2, and the one after that is n + 4. The left side of her equation is the sum of all three. The right side, 4n, is "four times the smallest." Her equation is a faithful translation of the sentence.
sum of the three integers = four times the smallest
Step 2 — Combine like terms on the left
There are three copies of n and the constants 2 and 4. Together that gives 3n and 6.
3n + 6 = 4n
Step 3 — Get the variable terms on one side
Both sides contain n, so subtract 3n from both sides. Subtracting the smaller coefficient keeps the remaining coefficient positive and avoids a negative sign.
6 = n
Step 4 — Find all three integers
The problem asks for the three integers, not just the smallest. Substitute n = 6 into each expression.
middle: n + 2 = 8
largest: n + 4 = 10
The integers are 6, 8 and 10, and 6 is even, so the starting assumption that n is even is satisfied.
Solution: Method 2 — Center the Integers on the Middle Value
Three consecutive even integers are symmetric around the middle one. If we call the middle integer m, the smallest is m − 2 and the largest is m + 2. Notice what happens when we add them:
The −2 and +2 cancel, so the sum of three consecutive integers is always three times the middle one. Now translate "four times the smallest" using the same variable: 4(m − 2).
3m = 4m − 8
8 = m
The middle integer is 8, so the other two are 8 − 2 = 6 and 8 + 2 = 10. Same answer, 6, 8, 10, from a different starting point.
Which is better? For this problem the two methods are about equally long, because the right side refers to the smallest integer, which is convenient in Method 1. When a problem refers to the sum or average, centering usually wins, because the constants cancel instantly.
Solution: Method 3 — Reason with the Offsets
This approach skips most of the algebra and asks: what is the difference between the two sides?
Each of the three integers is the smallest integer plus an offset: 0, 2 and 4. So the sum is 3n plus the offsets 0 + 2 + 4 = 6. Four times the smallest is 3n plus one more n. For the two quantities to be equal, the leftover pieces must match:
4 × smallest = 3n + n
so the extra pieces must agree: n = 6
In words: the smallest integer has to be exactly as large as the total of the "gaps" built into the other two numbers. That is a nice way to see why the answer is not an accident.
The three consecutive even integers are 6, 8 and 10, with n = 6 as the smallest.
Verification
Substitute back into Abigail's equation and into the original sentence. Both checks matter, since the first confirms the algebra and the second confirms the translation.
Check 1: Abigail's equation with n = 6.
Left side: 6 + (6+2) + (6+4) = 6 + 8 + 10 = 24
Right side: 4 × 6 = 24 ✓
Check 2: the words. Are 6, 8, 10 consecutive even integers? Each is even, and each is 2 more than the last ✓. Is their sum (24) four times the smallest (6)? 4 × 6 = 24 ✓
Check 3: Method 2's version. With middle 8: 3 × 8 = 24 and 4 × (8 − 2) = 24 ✓
Sanity Check: Is There Only One Answer, and Why This One?
The sum of three consecutive even integers is 3n + 6, which is always bigger than 3n but, for large n, much smaller than 4n. So the sum starts out ahead of "four times the smallest" for tiny numbers, then falls behind. They can only be equal at one crossover point. The table shows it:
| Smallest n | Integers | Sum | 4n | Sum vs. 4n |
|---|---|---|---|---|
| 2 | 2, 4, 6 | 12 | 8 | sum is bigger |
| 4 | 4, 6, 8 | 18 | 16 | sum is bigger |
| 6 | 6, 8, 10 | 24 | 24 | equal ✓ |
| 8 | 8, 10, 12 | 30 | 32 | 4n is bigger |
| 10 | 10, 12, 14 | 36 | 40 | 4n is bigger |
The gap closes by exactly 2 each time n increases by 2, and it hits zero at n = 6. A linear equation with different slopes on each side has exactly one solution, and this table lets you see that rather than just trust it.
Common Pitfalls
n + (n+1) + (n+2) = 4n gives 3n + 3 = 4n, so n = 3, and the "integers" 3, 4, 5 are not even. Adding 1 moves you from even to odd. Whenever the problem says even or odd, the step is 2. Only plain "consecutive integers" use a step of 1.
The question asks "What are the three integers?" and the value n = 6 is just one of them. Always reread the question after solving and write out the full list: 6, 8, 10.
From 3n + 6 = 4n, subtracting 4n from both sides gives 6 − n = 0, which is correct but easy to mishandle. Students sometimes write 6 = −n and then report n = −6. Then the "integers" are −6, −4, −2, whose sum is −12, while 4 × (−6) = −24. The check immediately exposes the error. Subtracting 3n instead keeps everything positive.
Writing 3n + 6 = 4(3n + 6) or 4(n + 4) attaches the "four times" to the wrong quantity. The phrase refers to one specific integer, the smallest, which is why the right side is just 4n. Abigail's equation gets this right.
How to Spot This Problem Type
- "Consecutive even/odd integers" signals a step of 2. Plain "consecutive integers" signals a step of 1.
- "Sum is … times the smallest/largest" signals an equation with the variable on both sides, so expect to collect terms.
- A named variable for the "smallest" signals Method 1, the standard setup. A problem about the sum or average hints at centering on the middle (Method 2).
- "Find the integers" means you need every number in the list, not just
n.
Disguised cousins include "three even house numbers in a row," "three consecutive even-numbered pages," and "three seats numbered consecutively on one side of a street" (house numbers and seat numbers on one side often skip by 2).
The Pattern Behind This
Any list of consecutive integers with step d starting at n has a simple sum formula for k terms:
sum = k·n + d·k(k−1)/2
For k = 3 and d = 2, that is 3n + 2·3 = 3n + 6, exactly what we found. Now suppose the problem says the sum is m times the smallest:
This tells us which multiplier values produce a valid answer. For m = 4, n = 6 (even ✓). For m = 6, n = 2 (even ✓). For m = 5, n = 3, which is a perfectly good integer but not even, so no set of three consecutive even integers works. For m = 3 there is no solution at all, since 3n + 6 = 3n is impossible.
Takeaway: solving the equation is only half the job. The result must also satisfy the type constraint (even, odd, positive, whole). When a problem is built from "nice" numbers, it is usually because the author chose the answer first. The multiplier 4 here was picked so that n = 6/(4−3) comes out even.
Four What-If Problems
These build in difficulty. Try each one on paper before opening the solution, and notice how much of Abigail's setup you can reuse.
Three consecutive odd integers have a sum that is five times the smallest of the three. What are the integers?
Let n be the smallest odd integer. Odd integers are also 2 apart, so the three are n, n + 2 and n + 4.
n + (n+2) + (n+4) = 5n, which simplifies to 3n + 6 = 5n.
Subtract 3n from both sides: 6 = 2n, so n = 3.
3, 3 + 2 = 5, 3 + 4 = 7. Note that 3 is odd, as required.
3 + 5 + 7 = 15 and 5 × 3 = 15 ✓. The integers are 3, 5, 7.
Four consecutive even integers have a sum that is five times the smallest of the four. What are the integers?
Let n be the smallest. The four integers are n, n + 2, n + 4 and n + 6.
n + (n+2) + (n+4) + (n+6) = 5n.
The left side is 4n + 12, so the equation becomes 4n + 12 = 5n.
Subtract 4n: n = 12. The integers are 12, 14, 16, 18.
12 + 14 + 16 + 18 = 60 and 5 × 12 = 60 ✓. The integers are 12, 14, 16, 18.
The sum of three consecutive even integers is 8 less than four times the smallest. What are the integers?
"8 less than" means subtract 8 after multiplying: 4n − 8. The phrase is written backwards compared with the algebra.
n + (n+2) + (n+4) = 4n − 8, so 3n + 6 = 4n − 8.
Subtract 3n from both sides: 6 = n − 8. Add 8 to both sides: n = 14.
14, 16, 18.
14 + 16 + 18 = 48 and 4 × 14 − 8 = 56 − 8 = 48 ✓. The integers are 14, 16, 18.
Three consecutive even integers have the property that the sum of the middle and largest integers is 10 more than three times the smallest. What are the integers?
Let n be the smallest. Then the middle is n + 2 and the largest is n + 4.
"Sum of the middle and largest" is (n+2) + (n+4) = 2n + 6. "10 more than three times the smallest" is 3n + 10. So 2n + 6 = 3n + 10.
Subtract 2n from both sides: 6 = n + 10. Subtract 10: n = −4.
−4, −4 + 2 = −2, −4 + 4 = 0. The answer includes negative integers and zero. Zero counts as an even integer, and nothing in the problem restricted us to positive numbers.
Middle + largest: −2 + 0 = −2. Three times smallest plus 10: 3(−4) + 10 = −12 + 10 = −2 ✓. The integers are −4, −2, 0.
Frequently Asked Questions
Let n be the smallest even integer. Consecutive even integers are 2 apart, so the next two are n + 2 and n + 4. In this example, the sum is n + (n + 2) + (n + 4) = 3n + 6. If that sum equals four times the smallest, 3n + 6 = 4n gives n = 6, so the integers are 6, 8 and 10.
Collect the variable terms on one side by subtracting the smaller variable term from both sides, then solve the remaining one-step equation. In this example, 3n + 6 = 4n becomes 6 = n after subtracting 3n from both sides. Always substitute back into the original statement to confirm: 6 + 8 + 10 = 24 and 4 × 6 = 24.
Even numbers alternate with odd numbers, so adding 1 to an even integer gives an odd integer. To land on the next even integer you must add 2. The same rule applies to consecutive odd integers. That is why the three integers in this problem are written n, n + 2 and n + 4, and why 6, 8, 10 is a valid answer while 6, 7, 8 would not be.
2026-11-01