AMC 10 · 2011 · #10
Grade 6 number-theoryA majority of the 30 students in Ms. Demeanor's class bought pencils at the school bookstore. Each of these students bought the same number of pencils, and this number was greater than 1. The cost of a pencil in cents was greater than the number of pencils each student bought, and the total cost of all the pencils was $$17.71$. What was the cost of a pencil in cents?
Pick an answer.
AMC 10 2011 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: In a class of $30$ students, more than half bought pencils. Every buyer bought the same number of pencils, more than $1$ each. The price of one pencil in cents was greater than the number of pencils each student bought, and all the pencils together cost $\$17.71$. Find the price of one pencil in cents.
Givens: There are $30$ students in the class; A majority of them bought pencils (more than half); Each buyer bought the same number of pencils, and that number is greater than $1$; The price of a pencil in cents is greater than the number of pencils each student bought; The total cost of all pencils is $\$17.71$; Answer choices: (A) $7$, (B) $11$, (C) $17$, (D) $23$, (E) $77$
Unknowns: The price of one pencil in cents
Understand
Restated: In a class of $30$ students, more than half bought pencils. Every buyer bought the same number of pencils, more than $1$ each. The price of one pencil in cents was greater than the number of pencils each student bought, and all the pencils together cost $\$17.71$. Find the price of one pencil in cents.
Givens: There are $30$ students in the class; A majority of them bought pencils (more than half); Each buyer bought the same number of pencils, and that number is greater than $1$; The price of a pencil in cents is greater than the number of pencils each student bought; The total cost of all pencils is $\$17.71$; Answer choices: (A) $7$, (B) $11$, (C) $17$, (D) $23$, (E) $77$
Plan
Primary tool: #4 Introduce a Variable
Secondary: #8 Analyze the Units, #7 Identify Subproblems, #3 Eliminate Possibilities
The total cost is (number of buyers) $\times$ (pencils each) $\times$ (price per pencil), so Tool #4 (Introduce a Variable) is the natural start: name the three unknowns and write one product equation, $s \cdot n \cdot c = 1771$. That turns the word problem into a number-factoring task. Tool #8 (Analyze the Units) first pins everything to one unit — cents — so the total is a whole-number product. Tool #7 (Identify Subproblems) handles the key sub-step — breaking $1771$ into its prime factors, which reveals exactly which whole numbers can appear. Tool #3 (Eliminate Possibilities) then uses the three inequalities (majority, pencils $>1$, price $>$ pencils) to lock each factor into its correct role and rule out the trap choices.
Execute — Answer: B
4.MD.A.2 Step 1 Put everything in cents
- The price is asked for in cents, so convert the total to cents to keep one unit throughout.
- Since $\$1 = 100$ cents, $\$17.71$ becomes $1771$ cents.
💡 Working in a single unit means every quantity is a whole number of cents, so the total is just a product of whole numbers.
6.EE.B.6 Step 2 Name the three unknowns
- Let $s$ be the number of students who bought pencils, $n$ the number of pencils each bought, and $c$ the price of one pencil in cents.
- The total cost is these three multiplied together, so $s \cdot n \cdot c = 1771$.
- The rules give $s > 15$ (a majority of $30$), $n > 1$, and $c > n$.
💡 One equation ties the three unknowns together; the inequalities are the extra clues that will separate them.
4.OA.B.4 Step 3 Factor the total
- Since $s$, $n$, and $c$ are whole numbers whose product is $1771$, each must be built from the prime factors of $1771$.
- Factoring gives $1771 = 7 \times 11 \times 23$.
- All three factors are prime, so $\{s, n, c\}$ can only be made from $7$, $11$, and $23$.
- Notice $17$ is not a factor of $1771$, so choice (C) is impossible right away.
💡 Prime factorization lists the only bricks available, so the three answers must be assembled from exactly $7$, $11$, and $23$.
6.EE.B.5 Step 4 Assign each factor with the clues
- There are three prime factors and three unknowns, so each unknown equals one of $7$, $11$, $23$.
- The number of buyers must be a majority of $30$, meaning greater than $15$: only $23$ qualifies, so $s = 23$.
- That leaves $n$ and $c$ as $7$ and $11$.
- Since the price must beat the pencils each ($c > n$), take $c = 11$ and $n = 7$.
- Check: $n = 7 > 1$ and $c = 11 > 7$, all rules hold.
- So one pencil costs $11$ cents, which is choice (B).
💡 Each inequality forces exactly one factor into its slot, so the roles are pinned down with no guessing left.
4.MD.A.2 The price is asked for in cents, so convert the total to cents to keep one unit 6.EE.B.6 Let $s$ be the number of students who bought pencils, $n$ the number of pencils 4.OA.B.4 Since $s$, $n$, and $c$ are whole numbers whose product is $1771$, each must be 6.EE.B.5 There are three prime factors and three unknowns, so each unknown equals one of Review
Reasonableness: Verify the full story with $s = 23$, $n = 7$, $c = 11$: $23 \times 7 \times 11 = 1771$ cents $= \$17.71$, matching the total. All conditions check out — $23$ is a majority of $30$, each student bought $7 > 1$ pencils, and each pencil cost $11 > 7$ cents. The price $11$ cents is choice $\textbf{(B)}$. Choice (C) $17$ was never possible since $17$ does not divide $1771$; (A) $7$ is the pencils-per-student count, not the price; (D) $23$ is the number of buyers; (E) $77$ is $7 \times 11$, not a single pencil's price.
Alternative: Skip naming variables and test the choices directly (Tool #3, Eliminate Possibilities): a valid price must divide $1771 = 7 \times 11 \times 23$. That kills (C) $17$ immediately. For the price to leave a majority-sized buyer count, dividing $1771$ by the price must give a product containing $23$. Only $11$ works: $1771 \div 11 = 161 = 23 \times 7$, giving $23$ buyers each with $7$ pencils and price $11 > 7$. So $\textbf{(B)}$.
CCSS standards used (min grade 6)
4.MD.A.2Use the four operations to solve word problems involving distances, time intervals, and money (Converting the total $\$17.71$ into $1771$ cents so every quantity is a whole number of cents.)6.EE.B.6Use variables to represent numbers and write expressions to solve problems (Naming the three unknowns $s$, $n$, $c$ and writing the product equation $s \cdot n \cdot c = 1771$.)4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Factoring $1771 = 7 \times 11 \times 23$ to find the only whole numbers that can multiply to the total.)6.EE.B.5Understand solving an equation or inequality as a process of finding values (Using the inequalities $s > 15$, $n > 1$, $c > n$ to decide which factor is the buyer count, the pencil count, and the price.)
⭐ When whole numbers multiply to a fixed total, factor the total into primes first — then the extra clues just tell you which factor plays which role.
⭐ When whole numbers multiply to a fixed total, factor the total into primes first — then the extra clues just tell you which factor plays which role.
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