AMC 10 · 2002 · #11
Grade 8 arithmeticThe product of three consecutive positive integers is 8 times their sum. What is the sum of their squares?
Pick an answer.
AMC 10 2002 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: Three whole numbers sit right next to each other, like $4,5,6$. When you multiply all three together, the result equals $8$ times what you get by adding them. Find the sum of the squares of those three numbers.
Givens: The three numbers are consecutive positive integers (each one more than the last); Their product equals $8$ times their sum; Answer choices: (A) $50$, (B) $77$, (C) $110$, (D) $149$, (E) $194$
Unknowns: The three integers themselves; The sum of their squares
Understand
Restated: Three whole numbers sit right next to each other, like $4,5,6$. When you multiply all three together, the result equals $8$ times what you get by adding them. Find the sum of the squares of those three numbers.
Givens: The three numbers are consecutive positive integers (each one more than the last); Their product equals $8$ times their sum; Answer choices: (A) $50$, (B) $77$, (C) $110$, (D) $149$, (E) $194$
Plan
Primary tool: #4 Introduce a Variable
Secondary: #13 Convert to Algebra, #3 Eliminate Possibilities
Naming the middle integer $n$ (Tool #4) is the smart move because the three consecutive numbers become $n-1$, $n$, $n+1$ — a symmetric setup that makes both the product and the sum collapse into clean expressions. Tool #13 (Convert to Algebra) then turns the sentence 'the product is $8$ times the sum' into one equation in $n$. That equation shrinks to $n^2=25$, and Tool #3 (Eliminate Possibilities) uses the word 'positive' to keep $n=5$ and discard $n=-5$.
Execute — Answer: B
6.EE.B.6 Step 1 Name the middle number
- Call the middle integer $n$.
- Because the numbers are consecutive, the one below is $n-1$ and the one above is $n+1$.
- Centering the name on the middle number is what makes the algebra symmetric and short.
💡 For evenly spaced numbers, naming the middle one lets the outer two balance around it.
6.EE.A.2 Step 2 Write the product and the sum
- The sum is $(n-1)+n+(n+1)=3n$ — the $-1$ and $+1$ cancel.
- The product is $(n-1)\,n\,(n+1)$.
- Multiplying the outer pair first, $(n-1)(n+1)=n^2-1$, so the product is $n(n^2-1)=n^3-n$.
💡 The symmetric $n-1$ and $n+1$ make the sum's ends cancel and the product fold into a difference of squares.
8.EE.A.2 Step 3 Solve for the middle number
- The condition 'product equals $8$ times the sum' becomes $n^3-n = 8\cdot 3n = 24n$.
- Move the $24n$ over: $n^3-n-24n=0$, so $n^3-25n=0$, which factors as $n(n^2-25)=0$.
- Since $n$ is a positive integer it is not $0$, so $n^2=25$, giving $n=5$ (the value $n=-5$ is discarded because the integers are positive).
💡 Dividing out the shared factor $n$ collapses a cubic into $n^2=25$, and 'positive' picks the single valid root.
6.EE.A.1 Step 4 Add up the squares
- With $n=5$ the three integers are $4,5,6$.
- Their squares are $4^2=16$, $5^2=25$, and $6^2=36$.
- Adding them gives $16+25+36=77$, which is choice (B).
💡 Once the numbers are pinned down, the answer is just three squares added together.
6.EE.B.6 Call the middle integer $n$. Because the numbers are consecutive, the one below 6.EE.A.2 The sum is $(n-1)+n+(n+1)=3n$ — the $-1$ and $+1$ cancel. The product is $(n-1)\ 8.EE.A.2 The condition 'product equals $8$ times the sum' becomes $n^3-n = 8\cdot 3n = 24 6.EE.A.1 With $n=5$ the three integers are $4,5,6$. Their squares are $4^2=16$, $5^2=25$, Review
Reasonableness: Check the numbers $4,5,6$ against the original wording: their product is $4\cdot5\cdot6=120$ and their sum is $4+5+6=15$; indeed $120=8\cdot15$, so the condition holds exactly. The sum of squares $16+25+36=77$ lands on choice (B), and it sits sensibly between the smaller options and the larger ones.
Alternative: Instead of algebra, guess and check: three consecutive numbers near the answer size. Try $4,5,6$: product $120$, sum $15$, and $120\div15=8$ — a match on the first sensible try, giving sum of squares $77$. The algebra just guarantees this is the only positive solution rather than a lucky guess.
CCSS standards used (min grade 8)
6.EE.B.6Use variables to represent numbers and write expressions to solve problems (Naming the middle integer $n$ so the three consecutive numbers become $n-1$, $n$, $n+1$.)6.EE.A.2Write, read, and evaluate expressions in which letters stand for numbers (Turning the sum into $3n$ and the product into $n^3-n$ using the symmetric names.)8.EE.A.2Use square root and cube root symbols to represent solutions (Reducing $n^3=25n$ to $n^2=25$ and taking the square root to get $n=5$.)6.EE.A.1Write and evaluate numerical expressions involving whole-number exponents (Squaring $4,5,6$ and adding $16+25+36=77$.)
⭐ For three consecutive numbers, name the middle one $n$: the sum becomes $3n$ and the product becomes $n^3-n$, and the equation shrinks to $n^2=25$.
⭐ For three consecutive numbers, name the middle one $n$: the sum becomes $3n$ and the product becomes $n^3-n$, and the equation shrinks to $n^2=25$.
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