AMC 8 · 2016 · #19
Easy mode Grade 6Problem
Picture a list of 25 even numbers in a row, where each one is 2 more than the one before. For example, a list like 2,4,6,8,… would follow this pattern, but the list here starts at a different number.
When you add up all 25 of these numbers, the total is 10,000.
What is the largest number in the list?
Pick an answer.
AMC 8 2016 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Take $25$ even integers in a row, like $..., n-4, n-2, n, n+2, n+4, ...$. Their sum is $10{,}000$. What is the largest one?
Givens: There are exactly $25$ integers; They are consecutive even integers (each is $2$ more than the previous); Their total sum is $10{,}000$; Answer choices: (A) $360$, (B) $388$, (C) $412$, (D) $416$, (E) $424$
Unknowns: The largest of the $25$ integers (the $25^{\text{th}}$ term)
Understand
Restated: Take $25$ even integers in a row, like $..., n-4, n-2, n, n+2, n+4, ...$. Their sum is $10{,}000$. What is the largest one?
Givens: There are exactly $25$ integers; They are consecutive even integers (each is $2$ more than the previous); Their total sum is $10{,}000$; Answer choices: (A) $360$, (B) $388$, (C) $412$, (D) $416$, (E) $424$
Plan
Primary tool: #5 Look for a Pattern
Secondary: #9 Solve an Easier Related Problem, #7 Identify Subproblems
Twenty-five consecutive even integers is too many to list, but the structure is highly regular. Tool #9 (Easier Problem) — try $3$ or $5$ consecutive evens first — exposes the pattern that the sum is always (count) $\times$ (middle term). Tool #5 (Pattern) lets us use that fact to jump straight to the middle term: $10{,}000 \div 25 = 400$. Then Tool #7 (Subproblems) splits the rest into two clean pieces — find the middle term, then step out $12$ places (by $+2$ each) to reach the largest. This dodges Tool #13 (Algebra) entirely.
Execute — Answer: E
3.OA.B.5 Step 1 - Try the easier version with $3$ consecutive evens.
- Take $4, 6, 8$: sum is $18$ and the middle term is $6$.
- Notice $18 = 3 \times 6$.
- Try $5$ consecutive evens $2, 4, 6, 8, 10$: sum is $30 = 5 \times 6$, and the middle term is $6$ again.
- The pattern: for an odd number of evenly-spaced terms, sum $=$ count $\times$ middle term.
💡 Working a small case before tackling $25$ terms shows why pairs above and below the middle cancel — a Grade 3 "properties of operations" idea.
6.SP.B.5 Step 2 - Apply the pattern to the real problem.
- The $25$ even integers have an odd count, so their sum equals $25 \times$ (middle term).
- Divide both sides by $25$ to recover the middle term.
💡 For an odd-length evenly-spaced list, the mean and the middle value are the same number — a Grade 6 statistics fact.
4.OA.C.5 Step 3 - Locate the middle position.
- With $25$ terms, the middle is the $13^{\text{th}}$ term (because $12$ sit to its left and $12$ to its right).
- So term $13$ equals $400$.
💡 Splitting the count into "left of middle", "middle", "right of middle" is a Grade 4 sequence-position move.
4.OA.C.5 Step 4 - Step from the $13^{\text{th}}$ term up to the $25^{\text{th}}$ term.
- That's $25 - 13 = 12$ steps, and each step adds $2$ because the integers are consecutive evens.
💡 Adding the common difference $12$ times to walk from term $13$ to term $25$ is the same skip-counting Grade 4 students do with patterns.
3.OA.B.5 Try the easier version with $3$ consecutive evens. Take $4, 6, 8$: sum is $18$ a 6.SP.B.5 Apply the pattern to the real problem. The $25$ even integers have an odd count, 4.OA.C.5 Locate the middle position. With $25$ terms, the middle is the $13^{\text{th}}$ 4.OA.C.5 Step from the $13^{\text{th}}$ term up to the $25^{\text{th}}$ term. That's $25 Review
Reasonableness: Sanity check the smallest term too: $400 - 12 \times 2 = 376$. So the list runs $376, 378, \ldots, 424$. Quick pairing: $(376 + 424) = 800$, and pairing every term with its mirror around the middle gives $12$ pairs of $800$ plus the lone middle $400$, total $12 \times 800 + 400 = 9600 + 400 = 10{,}000$. \checkmark\ The answer $424$ is the largest choice given and is consistent with the middle being a round $400$.
Alternative: Tool #3 (Eliminate Possibilities) on the choices: if the largest is $L$, the smallest is $L - 48$, and the sum is $25 \times \dfrac{L + (L-48)}{2} = 25(L - 24)$. Set $25(L - 24) = 10{,}000$ so $L - 24 = 400$ and $L = 424$. Only choice (E) works; (A)-(D) each fail this equation.
CCSS standards used (min grade 6)
3.OA.B.5Apply properties of operations as strategies to multiply and divide (Recognizing on the small cases ($3$ and $5$ terms) that the sum factors as count $\times$ middle term — a use of the distributive/commutative properties.)4.OA.C.5Generate a number or shape pattern that follows a given rule (Locating the $13^{\text{th}}$ term as the middle of $25$ and then stepping $12$ times by $+2$ to reach the $25^{\text{th}}$ term.)6.SP.B.5Summarize numerical data sets in relation to their context (mean, median) (Using the fact that for an evenly-spaced list with an odd number of terms, the mean ($10{,}000 / 25 = 400$) equals the middle value.)
⭐ If you have an odd number of evenly-spaced numbers, the average IS the middle one — a Grade 6 idea that cracks this AMC 8 problem in one division!
⭐ If you have an odd number of evenly-spaced numbers, the average IS the middle one — a Grade 6 idea that cracks this AMC 8 problem in one division!
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