AMC 8 · 2009 · #11

Grade 6 number-theory

Archetype Small-Domain Constrained Search └ Sub-archetype Factor / Divisibility Constraint

gcdprime-factorizationfactorsdivisibility-rules identify-subproblemssystematic-enumeration ↑ Prerequisites: multi-digit-arithmeticfactors

📏 Medium solution 💡 3 insights

Problem

At the Amaco Middle School bookstore, a pencil costs a whole number of cents. Some seventh graders bought one pencil each, paying

1.43</span> in total. Some of the <span class="mk-c">30 sixth graders</span> also bought one pencil each, paying <span class="mk-c">

1.95 in total. How many more sixth graders than seventh graders bought a pencil?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2009 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

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 $\frac{195 - 143}{c}$ 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.

1STEP 1

In cents the totals are $1.43 = 143 and$ 1.95 = 195, so the price c must divide both 143 and 195.

c ∣ 143 and c ∣ 195

2STEP 2

Factor each total to list its divisors: 143 = 11 × 13 and 195 = 3 × 5 × 13.

143 = 11 × 13, 195 = 3 × 5 × 13

3STEP 3

The only prime shared by the two factorizations is 13, so the common factors of 143 and 195 are just 1 and 13.

gcd(143, 195) = 13 → common factors = {1, 13}

4STEP 4

A 1-cent pencil needs $\frac{195}{1}$ = 195 buyers, but there are only 30 sixth graders, so c = 1 is out and c = 13.

c = 1 →

\frac{195}{1}

= 195 ＞ 30 sixth graders → reject. So c = 13

5STEP 5

Sixth graders paid 195 - 143 = 52 more cents, and each extra pencil is 13 cents, so the head-count gap is $\frac{52}{13}$ .

\frac{195 - 143}{13}

\frac{52}{13}

= 4 → (D)

Answer

Cross-check by counting each group directly. Sixth-grade buyers =

\frac{195}{13}

= 15 and seventh-grade buyers =

\frac{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.

💡Key takeaway

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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