AMC 8 · 2012 · #13
Easy mode Grade 6Problem
A school bookstore sells pencils. Every pencil costs the same whole number of cents, and each pencil costs more than cent.
Jamar buys some of these pencils and pays exactly . Sharona buys some of the same pencils and pays exactly .
How many more pencils did Sharona buy than Jamar?
Pick an answer.
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.