AMC 8 · 2009 · #11
Grade 6 number-theoryProblem
The Amaco Middle School bookstore sells pencils costing a whole number of cents. Some seventh graders each bought a pencil, paying a total of 1.43 dollars. Some of the 30 sixth graders each bought a pencil, and they paid a total of 1.95 dollars. How many more sixth graders than seventh graders bought a pencil?
Pick an answer.
AMC 8 2009 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 Amaco Middle School bookstore, a pencil costs a whole number of cents. Some seventh graders bought one pencil each, paying $\$1.43$ in total. Some of the $30$ sixth graders also bought one pencil each, paying $\$1.95$ in total. How many more sixth graders than seventh graders bought a pencil?
Givens: Pencil cost is a whole number of cents; Total seventh-grade spending $= \$1.43 = 143$ cents; Total sixth-grade spending $= \$1.95 = 195$ cents; There are only $30$ sixth graders, so the number of sixth-grade buyers is at most $30$; Answer choices: (A) $1$, (B) $2$, (C) $3$, (D) $4$, (E) $5$
Unknowns: (sixth-grade buyers) $-$ (seventh-grade buyers)
Understand
Restated: At the Amaco Middle School bookstore, a pencil costs a whole number of cents. Some seventh graders bought one pencil each, paying $\$1.43$ in total. Some of the $30$ sixth graders also bought one pencil each, paying $\$1.95$ in total. How many more sixth graders than seventh graders bought a pencil?
Givens: Pencil cost is a whole number of cents; Total seventh-grade spending $= \$1.43 = 143$ cents; Total sixth-grade spending $= \$1.95 = 195$ cents; There are only $30$ sixth graders, so the number of sixth-grade buyers is at most $30$; Answer choices: (A) $1$, (B) $2$, (C) $3$, (D) $4$, (E) $5$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #2 Make a Systematic List, #3 Eliminate Possibilities
The question asks for a difference in head counts, but we cannot count heads until we know the pencil price. Tool #7 (Identify Subproblems) splits the work cleanly: (1) pin down the pencil cost $c$, then (2) compute $(195 - 143) / c$ for the difference. Subproblem (1) is a number-theory hunt: $c$ must divide both $143$ and $195$, so we list the factors of each (Tool #2) and keep only the common ones. Tool #3 (Eliminate Possibilities) then rules out the bad common factors — $c = 1$ would force $195$ sixth-grade buyers in a class of only $30$ — leaving a unique price.
Execute — Answer: D
4.OA.A.3 Step 1 - Convert dollars to cents so the pencil price is an ordinary whole number.
- $\$1.43 = 143$ cents and $\$1.95 = 195$ cents.
- Both totals must be exact multiples of the pencil price $c$, so $c$ is a common factor of $143$ and $195$.
💡 Naming the two subproblems — "which $c$ is allowed" and "how many buyers" — is the Tool #7 move that turns a vague word problem into a two-step calculation.
4.OA.B.4 Step 2 - Factor each total to list its divisors.
- Try small primes on $143$: $143 / 11 = 13$, so $143 = 11 \times 13$.
- For $195$: $195 / 5 = 39 = 3 \times 13$, so $195 = 3 \times 5 \times 13$.
💡 Writing out the prime factorization is the Tool #2 systematic list of factors, and recognizing $11$ and $13$ as primes is the Grade 4 factor/multiple skill.
6.NS.B.4 Step 3 - Take the greatest common factor.
- The only prime shared by $143 = 11 \times 13$ and $195 = 3 \times 5 \times 13$ is $13$, so the common factors of $143$ and $195$ are exactly $1$ and $13$.
💡 The GCF reading off prime factorizations is the Grade 6 number-theory move that turns the systematic list into a short candidate set.
4.OA.A.3 Step 4 - Eliminate $c = 1$.
- If a pencil cost $1$ cent, the sixth graders would have bought $195 / 1 = 195$ pencils — but there are only $30$ sixth graders total.
- So $c = 1$ is impossible, leaving $c = 13$ cents.
💡 Crossing off the candidate that breaks the "$\leq 30$ sixth graders" constraint is the Tool #3 elimination step.
4.NBT.B.6 Step 5 - Now solve subproblem (2) using $c = 13$.
- The sixth graders paid $195 - 143 = 52$ more cents than the seventh graders, and every extra pencil costs $13$ cents, so the extra head count is $52 / 13$.
💡 Subtracting the totals first (instead of counting each group separately) is the cleanest Tool #7 path; the final $52 \div 13 = 4$ is a Grade 4 whole-number division.
4.OA.A.3 Convert dollars to cents so the pencil price is an ordinary whole number. $\$1.4 4.OA.B.4 Factor each total to list its divisors. Try small primes on $143$: $143 / 11 = 1 6.NS.B.4 Take the greatest common factor. The only prime shared by $143 = 11 \times 13$ a 4.OA.A.3 Eliminate $c = 1$. If a pencil cost $1$ cent, the sixth graders would have bough 4.NBT.B.6 Now solve subproblem (2) using $c = 13$. The sixth graders paid $195 - 143 = 52$ Review
Reasonableness: Cross-check by counting each group directly. Sixth-grade buyers $= 195 / 13 = 15$ and seventh-grade buyers $= 143 / 13 = 11$. Difference $= 15 - 11 = 4$, matching choice (D). The sixth-grade count $15$ is well under the $30$-student cap, confirming $c = 13$ is the right pencil price.
Alternative: Tool #6 (Guess and Check) on the choices: for each candidate difference $d$, the sixth graders bought $d$ more pencils than the seventh graders, so the pencil price $c$ must equal $(195 - 143) / d = 52 / d$. Try $d = 1, 2, 3, 4, 5$: only $d = 1, 2, 4$ make $52/d$ a whole number, giving $c = 52, 26, 13$. Of those, $c = 52$ fails ($195/52$ is not whole) and $c = 26$ fails ($143/26$ is not whole), so only $c = 13$ survives — forcing $d = 4$, i.e. (D).
CCSS standards used (min grade 6)
4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Translating the dollar totals into whole-cent constraints ($c \mid 143$ and $c \mid 195$) and eliminating $c = 1$ via the $\leq 30$ sixth-grade-class size.)4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Factoring $143 = 11 \times 13$ and $195 = 3 \times 5 \times 13$ by trying small prime divisors and recognizing $11$ and $13$ as primes.)6.NS.B.4Find greatest common factor and least common multiple of two numbers (Reading $\gcd(143, 195) = 13$ off the prime factorizations and listing the common factors $\{1, 13\}$ as the candidate pencil prices.)4.NBT.B.6Find whole-number quotients and remainders with up to four-digit dividends (Performing the final division $52 \div 13 = 4$ (and the cross-check divisions $195 \div 13 = 15$, $143 \div 13 = 11$).)
⭐ Once you spot that the pencil price is the greatest common factor of $143$ and $195$, this AMC 8 problem reduces to a Grade 6 GCF on top of plain Grade 4 division.
⭐ Once you spot that the pencil price is the greatest common factor of $143$ and $195$, this AMC 8 problem reduces to a Grade 6 GCF on top of plain Grade 4 division.
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