AMC 10 · 2011 · #10

Grade 6 number-theorylogic
Archetype Small-Domain Constrained Search └ Sub-archetype Factor-Pair Enumeration
prime-factorizationfactorslinear-diophantine convert-to-algebracasework ↑ Prerequisites: prime-factorization
📏 Medium solution 💡 3 insights
Problem
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.

Pick an answer.

(A)
7
(B)
11
(C)
17
(D)
23
(E)
77

AMC 10 2011 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Introduce a Variable

The total cost is (number of buyers) × (pencils each) × (price per pencil), so Tool #4 (Introduce a Variable) is the natural start: name the three unknowns and write one product equation, s · n · 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.

1STEP 1

Put everything in cents

The price is asked in cents, so convert the total first: $17.71 becomes 1771 cents.

$17.71 = 17.71 × 100 = 1771 cents
2STEP 2

Name the three unknowns

Let s be the buyers, n the pencils each bought, c the price in cents; then s·n·c = 1771.

s · n · c = 1771, s ＞ 15, n ＞ 1, c ＞ n
3STEP 3

Factor the total

1771 factors as 7 × 11 × 23, all prime — so those are the only numbers available, and 17 is not one.

1771 = 7 × 11 × 23 (7, 11, 23 all prime)
4STEP 4

Assign each factor with the clues

Only 23 beats 15, so s = 23; the price must top the count, so c = 11 and n = 7.

s = 23 ( ＞ 15); n · c = 77 = 7 × 11, c ＞ n → n = 7, c = 11 → (B) 11
Answer
11
Verify the full story with s = 23, n = 7, c = 11: 23 × 7 × 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 (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 × 11, not a single pencil's price.
💡Key takeaway

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.

  • Put everything in cents
  • Name the three unknowns
  • Factor the total
  • Assign each factor with the clues

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