AMC 10 · 2006 · #9

Grade 7 number-theory
Archetype Small-Domain Constrained Search └ Sub-archetype Factor / Divisibility Constraint
linear-diophantineinterval-arithmeticlogical-deduction convert-to-algebrabound-inequality-then-enumerate ↑ Prerequisites: linear-diophantine
📏 Medium solution 💡 2 insights
Problem
Thirteen pencils and three erasers cost exactly 100 cents, with whole-cent prices. A pencil costs strictly more than an eraser. Find the cost of one pencil plus one eraser.

Pick an answer.

(A)
10
(B)
12
(C)
15
(D)
18
(E)
20
How to solve
Strategy Extreme Principle

Tool #13 (Convert to Algebra) and Tool #4 (Introduce a Variable) turn the shopping sentence into one equation in two unknowns: 13p + 3e = 100. One equation with two unknowns normally has many solutions, so the real work is done by the two side conditions — whole cents, and pencil dearer than eraser. Tool #14 (Extreme Principle) is the sharpest way to use them: push the eraser price to its cheapest allowed value to cap the pencil price from above, then push it to its dearest allowed value to floor the pencil price from below. The two bounds close on a single integer, so no case list is needed at all. Tool #3 (Eliminate Possibilities) then serves as a check rather than as the method, and explains why one wrong choice is on the list.

1STEP 1

Turn the purchase into an equation

The purchase becomes one equation plus two whole-number conditions.

13p + 3e = 100, p ＞ e ≥ 1, p, e ∈ Z
2STEP 2

Cheapest eraser caps the pencil

The cheapest possible eraser caps the pencil at 7.

e ≥ 1 → 13p = 100 - 3e ≤ 97 → p ≤ 97/13 ＜ 7.5 → p ≤ 7
3STEP 3

Pencil dearer than eraser floors it

The strict inequality pushes it back up to 7, pinning it exactly.

e ＜ p → 100 = 13p + 3e ＜ 16p → p ＞ 100/16 = 6.25 → p ≥ 7; with p ≤ 7, p = 7
4STEP 4

Solve for the eraser and check it

Solving and checking every condition gives the sum 10, choice (A).

13(7) = 91, 3e = 100 - 91 = 9, e = 3; p + e = 7 + 3 = 10 → (A)
Answer
10
The pair (p, e) = (7, 3) passes every condition at once: 13 × 7 + 3 × 3 = 91 + 9 = 100 cents exactly, both prices are whole cents, and the pencil at 7 beats the eraser at 3. A quick feel for the size also fits — 16 items sharing a dollar averages 100/16 = 6.25 cents each, and since pencils are the majority and the dearer item, the pencil should sit just above that average and the eraser well below, which is exactly what 7 and 3 do. The other choices die with the price: 10 is the only sum reachable, so (B), (C) and (D) have no price pair behind them at all.
💡Key takeaway

When one equation has two unknowns, squeeze the answer between a highest-possible and a lowest-possible value — if only one whole number survives, you are done.

  • Turn the purchase into an equation
  • Cheapest eraser caps the pencil
  • Pencil dearer than eraser floors it
  • Solve for the eraser and check it

More like this

Same archetype — closest grade level first.