AMC 8 · 2013 · #17
Easy mode Grade 6Problem
Imagine six positive whole numbers in a row, where each one is exactly 1 more than the one before it. So they look like n,n+1,n+2,n+3,n+4,n+5 for some starting number n.
When you add all six of these numbers together, the total is 2013.
What is the largest of the six numbers?
Pick an answer.
AMC 8 2013 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: Six positive integers in a row (each one bigger than the last) add up to $2013$. Find the largest of these six numbers.
Givens: Six numbers that are consecutive positive integers (e.g. $n, n+1, n+2, n+3, n+4, n+5$); Their sum equals $2013$; Answer choices: (A) $335$, (B) $338$, (C) $340$, (D) $345$, (E) $350$
Unknowns: The largest of the six consecutive integers
Understand
Restated: Six positive integers in a row (each one bigger than the last) add up to $2013$. Find the largest of these six numbers.
Givens: Six numbers that are consecutive positive integers (e.g. $n, n+1, n+2, n+3, n+4, n+5$); Their sum equals $2013$; Answer choices: (A) $335$, (B) $338$, (C) $340$, (D) $345$, (E) $350$
Plan
Primary tool: #5 Look for a Pattern
Secondary: #7 Identify Subproblems
Six consecutive integers have a clean pattern: their sum is always $6$ times the average, and the average sits exactly in the middle of the list. Tool #5 (Look for a Pattern) lets us turn the sum directly into an average without algebra. Tool #7 (Identify Subproblems) splits the work into two easy pieces — first find the average, then count up to the largest. We avoid Tool #13 (Algebra) because the pattern lets a young solver finish using only division and a small addition.
Execute — Answer: B
6.SP.B.5 Step 1 - Notice the pattern that makes consecutive sums easy.
- For any list of consecutive integers, $\text{sum} = (\text{count}) \times (\text{average})$, so $\text{average} = \text{sum} \div \text{count}$.
- With $6$ numbers summing to $2013$, divide to get the average.
💡 Treating the sum as $6 \times \text{average}$ is the Grade 6 idea that the mean is the "balance point" of the data.
5.NBT.B.7 Step 2 - Locate the average inside the list.
- With an even count ($6$) of consecutive integers, the average falls halfway between the two middle numbers (the $3$rd and $4$th).
- So $335.5$ means the $3$rd number is $335$ and the $4$th number is $336$.
💡 A decimal like $.5$ landing between two whole numbers is the Grade 5 "halfway point" reading of a decimal.
3.OA.A.3 Step 3 - Break the goal into a subproblem: once the $3$rd number is known, just count up.
- The largest is the $6$th number, which is $3$ steps after the $3$rd number.
💡 Adding $3$ to the $3$rd term to reach the $6$th term is a Grade 3 multi-step word-problem add.
3.OA.A.3 Step 4 Match the largest, $338$, to the answer choices to finish the problem.
💡 Closing the final subproblem by comparing your value to the listed options is a Grade 3 multi-step move.
6.SP.B.5 Notice the pattern that makes consecutive sums easy. For any list of consecutive 5.NBT.B.7 Locate the average inside the list. With an even count ($6$) of consecutive inte 3.OA.A.3 Break the goal into a subproblem: once the $3$rd number is known, just count up. 3.OA.A.3 Match the largest, $338$, to the answer choices to finish the problem. Review
Reasonableness: Write the six numbers out and add: $333 + 334 + 335 + 336 + 337 + 338$. Pair the ends: $333 + 338 = 671$, $334 + 337 = 671$, $335 + 336 = 671$. Three pairs of $671$ give $3 \times 671 = 2013$. The sum matches, and the largest is $338$ — answer (B).
Alternative: Tool #6 (Guess and Check) on the choices. If the largest were $335$ (A), the six numbers would be $330$–$335$, summing to $6 \times 332.5 = 1995$ — too small. If the largest were $340$ (C), the sum would be $6 \times 337.5 = 2025$ — too big. Only (B) $338$ gives $6 \times 335.5 = 2013$. Tool #3 (Eliminate) confirms (B) is the only choice that works.
CCSS standards used (min grade 6)
3.OA.A.3Use multiplication and division within 100 to solve word problems (The closing steps — adding $335 + 3 = 338$ to reach the $6$th term and matching the value to a multiple-choice option — are Grade 3 multi-step whole-number reasoning.)5.NBT.B.7Add, subtract, multiply, and divide decimals to hundredths (Reading $335.5$ as the halfway point between $335$ and $336$ uses Grade 5 fluency with decimals to the tenths place.)6.SP.B.5Summarize numerical data sets, including the mean (average) of a distribution (Using $\text{sum} = \text{count} \times \text{average}$, so $\text{average} = 2013 \div 6 = 335.5$, is the Grade 6 mean concept applied to a six-number list.)
⭐ Six numbers in a row add up to $6$ times their average — once you find the average ($2013 \div 6 = 335.5$), the largest is just three steps past the middle.
⭐ Six numbers in a row add up to $6$ times their average — once you find the average ($2013 \div 6 = 335.5$), the largest is just three steps past the middle.
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