AMC 10 · 2006 · #9
Grade 6 arithmeticHow many sets of two or more consecutive positive integers have a sum of 15?
Pick an answer.
AMC 10 2006 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: Count how many different runs of two or more consecutive positive integers (like $4,5,6$) add up to exactly $15$.
Givens: The integers in each set are consecutive (each one is $1$ more than the previous); Every integer used must be a positive whole number ($1, 2, 3, \dots$); Each set must contain at least $2$ numbers; The numbers in a set must sum to $15$; Answer choices: (A) $1$, (B) $2$, (C) $3$, (D) $4$, (E) $5$
Unknowns: How many such sets of consecutive positive integers exist
Understand
Restated: Count how many different runs of two or more consecutive positive integers (like $4,5,6$) add up to exactly $15$.
Givens: The integers in each set are consecutive (each one is $1$ more than the previous); Every integer used must be a positive whole number ($1, 2, 3, \dots$); Each set must contain at least $2$ numbers; The numbers in a set must sum to $15$; Answer choices: (A) $1$, (B) $2$, (C) $3$, (D) $4$, (E) $5$
Plan
Primary tool: #4 Introduce a Variable
Secondary: #14 Extreme Principle, #2 Make a Systematic List
A set of consecutive integers is pinned down by two numbers: where it starts and how many terms it has. Tool #4 (Introduce a Variable) names those as $a$ and $n$ and turns the sum into one clean equation, $n(2a+n-1)=30$. Tool #14 (Extreme Principle) then caps how long a run can be — the shortest possible run of $n$ positive integers already sums to $1+2+\cdots+n$, which cannot exceed $15$ — so only a handful of lengths are even possible. Tool #2 (Make a Systematic List) checks those few lengths one by one and counts the winners.
Execute — Answer: C
6.EE.A.2 Step 1 Name the start and the length
- Let a set start at the positive integer $a$ and have $n$ terms, so it is $a, a+1, \dots, a+(n-1)$.
- Its sum is $n$ copies of $a$ plus the extra $0+1+\cdots+(n-1) = \tfrac{n(n-1)}{2}$.
- Setting that equal to $15$ and doubling both sides clears the fraction into a tidy product equation.
💡 A run of consecutive numbers is fixed by where it starts and how many there are, so give both a name.
4.OA.B.4 Step 2 Cap how long the run can be
- Because every term is at least $1$, the second factor $2a+n-1$ is at least $n+1$, which is bigger than $n$ itself.
- So in the product $n(2a+n-1)=30$, the length $n$ is the smaller factor, meaning $n \times n < 30$.
- Thus $n \le 5$, and since a set needs at least $2$ numbers, only $n = 2, 3, 4, 5$ are worth testing.
💡 The starting number drags the total up fast, so a run that sums to only $15$ cannot be very long.
6.EE.B.7 Step 3 Test each possible length
- For each candidate $n$, the other factor must equal $\tfrac{30}{n}$, and then $a = \tfrac{1}{2}\left(\tfrac{30}{n} - n + 1\right)$ must come out to a positive whole number.
- $n=2$: $\tfrac{30}{2}=15$, giving $a=7$, the set $\{7,8\}$.
- $n=3$: $\tfrac{30}{3}=10$, giving $a=4$, the set $\{4,5,6\}$.
- $n=4$: $\tfrac{30}{4}=7.5$ is not a whole number, so no set.
- $n=5$: $\tfrac{30}{5}=6$, giving $a=1$, the set $\{1,2,3,4,5\}$.
💡 Only a length that divides $30$ evenly and leaves a positive start can actually work.
4.OA.B.4 Step 4 Count the winning sets
- Three lengths produced a valid set of consecutive positive integers — $\{7,8\}$, $\{4,5,6\}$, and $\{1,2,3,4,5\}$ — while $n=4$ failed.
- That is $3$ sets in total, which is choice (C).
💡 Each length gives at most one run, so just tally the ones that survived.
6.EE.A.2 Let a set start at the positive integer $a$ and have $n$ terms, so it is $a, a+1 4.OA.B.4 Because every term is at least $1$, the second factor $2a+n-1$ is at least $n+1$ 6.EE.B.7 For each candidate $n$, the other factor must equal $\tfrac{30}{n}$, and then $a 4.OA.B.4 Three lengths produced a valid set of consecutive positive integers — $\{7,8\}$, Review
Reasonableness: Each found set really does sum to $15$: $7+8=15$, $4+5+6=15$, and $1+2+3+4+5=15$. The bound $n \le 5$ is safe because a run of $6$ smallest positive integers already sums to $1+2+3+4+5+6=21 > 15$, so nothing was missed above $5$. Three fits comfortably in the middle of the answer choices, not at an extreme.
Alternative: Use the average instead of algebra. A run of consecutive integers has a nice average that lands exactly in its middle, and (sum) = (average)$\times$(count), so the average must be $\tfrac{15}{n}$. An odd-length run has a whole-number middle, forcing $n$ to divide $15$: $n=3$ gives middle $5$, i.e. $\{4,5,6\}$, and $n=5$ gives middle $3$, i.e. $\{1,2,3,4,5\}$ ($n=15$ is too long). An even-length run has a half-number average, so $\tfrac{15}{n}$ must end in $.5$; only $n=2$ works, giving $\{7,8\}$. Again $3$ sets, choice (C).
CCSS standards used (min grade 6)
6.EE.A.2Write, read, and evaluate expressions in which letters stand for numbers (Naming the start $a$ and length $n$ and writing the sum as $n(2a+n-1)=30$.)4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Treating the length as the smaller factor of $30$ to bound $n$, and tallying which lengths divide $30$ evenly.)6.EE.B.7Solve real-world problems by writing and solving equations of the form px = q (Solving $2a+n-1=\tfrac{30}{n}$ for the starting value $a$ at each candidate length.)
⭐ Describe a run of consecutive numbers by where it starts and how many there are, turn the sum into $n(2a+n-1)=30$, and only a short run can add up to just $15$.
⭐ Describe a run of consecutive numbers by where it starts and how many there are, turn the sum into $n(2a+n-1)=30$, and only a short run can add up to just $15$.
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