AMC 8 · 2012 · #13
Grade 6 number-theoryProblem
Jamar bought some pencils costing more than a penny each at the school bookstore and paid \textdollar1.43. Sharona bought some of the same pencils and paid \textdollar1.87. How many more pencils did Sharona buy than Jamar?
Pick an answer.
AMC 8 2012 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: At the same school bookstore, Jamar paid $\$1.43$ for some pencils and Sharona paid $\$1.87$ for some of the same pencils. Each pencil costs a whole number of cents that is more than $1$. How many more pencils did Sharona buy than Jamar?
Givens: Jamar's total = $\$1.43 = 143$ cents; Sharona's total = $\$1.87 = 187$ cents; Both bought the same kind of pencil (one fixed whole-cent price); Each pencil costs more than $1$ cent; Answer choices: (A) $2$, (B) $3$, (C) $4$, (D) $5$, (E) $6$
Unknowns: The difference between Sharona's pencil count and Jamar's pencil count
Understand
Restated: At the same school bookstore, Jamar paid $\$1.43$ for some pencils and Sharona paid $\$1.87$ for some of the same pencils. Each pencil costs a whole number of cents that is more than $1$. How many more pencils did Sharona buy than Jamar?
Givens: Jamar's total = $\$1.43 = 143$ cents; Sharona's total = $\$1.87 = 187$ cents; Both bought the same kind of pencil (one fixed whole-cent price); Each pencil costs more than $1$ cent; Answer choices: (A) $2$, (B) $3$, (C) $4$, (D) $5$, (E) $6$
Plan
Primary tool: #6 Guess and Check
Secondary: #13 Convert to Algebra, #7 Identify Subproblems
The price per pencil $c$ has to divide both $143$ and $187$, and the problem says $c > 1$. With Tool #13 (Convert to Algebra) we write the two simple equations $j \cdot c = 143$ and $s \cdot c = 187$, which makes "$c$ is a common divisor" obvious. Then Tool #6 (Guess and Check) is the lightest way to find $c$ — try small primes ($2, 3, 5, 7, 11, \ldots$) on $143$ until one fits, and confirm it also divides $187$. Tool #7 (Identify Subproblems) splits the work into three clean steps: find $c$, then find $j$ and $s$, then subtract.
Execute — Answer: C
6.EE.B.6 Step 1 - Convert both totals to cents so every quantity is a whole number, and name the unknowns.
- Let $c$ be the price of one pencil in cents, $j$ Jamar's count, $s$ Sharona's count.
💡 Naming the unknown with a letter and writing the two "count $\times$ price = total" equations is the Grade 6 "use variables to represent numbers" move.
4.OA.B.4 Step 2 - Because the equations $j \cdot c = 143$ and $s \cdot c = 187$ both have whole-number solutions, $c$ must be a common divisor of $143$ and $187$.
- Find the divisors of $143$ by guess-and-check on small primes.
💡 Testing $2, 3, 5, 7, 11, \ldots$ until one works is the standard Grade 4 "find factor pairs" routine.
6.NS.B.4 Step 3 - Now check whether the same prime $11$ divides $187$.
- If it does, $11$ is a common divisor bigger than $1$, which is exactly the kind of $c$ we need.
💡 Confirming a common factor of two numbers is exactly the Grade 6 "greatest common factor" idea.
6.NS.B.4 Step 4 - Could $c$ be anything other than $11$?
- The divisors of $143$ are $\{1, 11, 13, 143\}$ and the divisors of $187$ are $\{1, 11, 17, 187\}$.
- Their common divisors are $1$ and $11$, and $c > 1$ rules out $1$.
- So $c = 11$ cents is forced.
💡 Once we have full factor lists, comparing them is a checking step, not a leap of faith.
4.NBT.B.6 Step 5 Plug $c = 11$ back into each equation to find how many pencils each kid bought, then subtract to answer the question.
💡 Dividing a 3-digit number by an 11 is a Grade 4 multi-digit division, and the subtraction is the last subproblem of the split.
6.EE.B.6 Convert both totals to cents so every quantity is a whole number, and name the u 4.OA.B.4 Because the equations $j \cdot c = 143$ and $s \cdot c = 187$ both have whole-nu 6.NS.B.4 Now check whether the same prime $11$ divides $187$. If it does, $11$ is a commo 6.NS.B.4 Could $c$ be anything other than $11$? The divisors of $143$ are ${1, 11, 13, 1 4.NBT.B.6 Plug $c = 11$ back into each equation to find how many pencils each kid bought, Review
Reasonableness: Sanity check the units and the size of the answer. At $11$ cents each, $13$ pencils cost $13 \times 11 = 143$ cents $= \$1.43$ and $17$ pencils cost $17 \times 11 = 187$ cents $= \$1.87$ — both totals match exactly. The difference $17 - 13 = 4$ is a small positive integer, matching one of the answer choices. If we had picked $c = 1$ (a penny), the count difference would be $187 - 143 = 44$, which isn't an option — confirming the problem's $c > 1$ hint really is the key constraint.
Alternative: Tool #3 (Eliminate Possibilities) on the answer choices. If $s - j = k$, then $sc - jc = kc$, so $kc = 187 - 143 = 44$. That means $c$ must equal $44 / k$ for the answer $k$. Try each choice: $k = 2 \Rightarrow c = 22$, but $22$ does not divide $143$; $k = 3 \Rightarrow c = 44/3$, not a whole number; $k = 4 \Rightarrow c = 11$, and $11$ divides both $143$ and $187$ — this works; $k = 5 \Rightarrow c = 8.8$, not whole; $k = 6 \Rightarrow c = 44/6$, not whole. Only $k = 4$ survives, so the answer is $\textbf{(C)}$.
CCSS standards used (min grade 6)
4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Testing small primes to factor $143$ as $11 \times 13$ — the Grade 4 factor-pair routine.)4.NBT.B.6Find whole-number quotients and remainders with up to four-digit dividends (Computing $143 \div 11 = 13$ and $187 \div 11 = 17$ to get the pencil counts.)6.NS.B.4Find greatest common factor and least common multiple of two numbers (Recognizing $c$ as a common divisor of $143$ and $187$ and identifying $11$ (in fact $\gcd(143, 187) = 11$) as the only choice with $c > 1$.)6.EE.B.6Use variables to represent numbers and write expressions to solve problems (Letting $c$, $j$, $s$ stand for the price and the two pencil counts, and writing $j \cdot c = 143$ and $s \cdot c = 187$.)
⭐ This AMC 8 problem just needs Grade 6 GCF thinking: find the only price bigger than $1$ cent that divides both totals, then divide to get the counts.
⭐ This AMC 8 problem just needs Grade 6 GCF thinking: find the only price bigger than $1$ cent that divides both totals, then divide to get the counts.
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