AMC 10 · 2004 · #13

Grade 6 arithmetic
Archetype Small-Domain Constrained Search └ Sub-archetype Factor / Divisibility Constraint
decimal-arithmeticlinear-diophantinedivisibility-rules bound-inequality-then-enumeratesystematic-enumeration ↑ Prerequisites: decimal-arithmetic
📏 Medium solution 💡 3 insights
Problem
Coins come in four thicknesses: penny 1.55 mm, nickel 1.95 mm, dime 1.35 mm, quarter 1.75 mm. Coins of these four types, each type usable any number of times, are stacked so that the pile is exactly 14 mm tall. Find how many coins are in the stack.

Pick an answer.

(A)
7
(B)
8
(C)
9
(D)
10
(E)
11

AMC 10 2004 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Eliminate Possibilities

The answer choices are the possible coin counts themselves (7 through 11), so the job is to knock out the impossible counts until one survives — the signature of Tool #3 (Eliminate Possibilities). Two filters do the knocking. First, Tool #8 (Analyze the Units): measure each coin against a dime and notice every coin is a dime plus a whole number of 0.20 mm steps, which forces 14 - 1.35n to be a multiple of 0.20. Second, Tool #14 (Extreme Principle): the thinnest and thickest coins pin the count between 8 and 10. Together the filters leave exactly one count standing.

1STEP 1

Measure each coin above a dime

Above the thinnest dime (1.35), penny, quarter and nickel sit 0.20, 0.40, 0.60 up, so any height is 1.35n plus a multiple of 0.20.

0.20, 0.40, 0.60 = 1 × 0.20, 2 × 0.20, 3 × 0.20 → height = 1.35n + 0.20×(whole number)
2STEP 2

Trap the count between 8 and 10

All nickels: 1.95 × 7 = 13.65 ＜ 14; all dimes: 1.35 × 11 = 14.85 ＞ 14. So n is 8, 9, or 10.

1.95 × 7 = 13.65 ＜ 14 ≤ 1.95n → n ≥ 8; 1.35n ≤ 14 ＜ 14.85 = 1.35 × 11 → n ≤ 10
3STEP 3

Keep only counts that fit the 0.20 step

14 - 1.35n must be a multiple of 0.20: n=8 → 3.20 = 0.20 × 16 ✓, n=9 → 1.85 ✗, n=10 → 0.50 ✗.

n=8: 3.20=0.20 × 16 ✓ n=9: 1.85=0.20 × 9.25 × n=10: 0.50=0.20 × 2.5 ×
4STEP 4

Confirm a real stack of 8 exists

Eight quarters really stack to 8 × 1.75 = 14.00 mm exactly, so this count is built, not just left over. Answer (B).

8 × 1.75 = 14.00 mm → n = 8 (B)
Answer
8
Eight quarters give 8 × 1.75 = 14 mm on the nose, so the count 8 is genuinely achievable, not just left over after elimination. The rejected neighbors fail cleanly: 7 coins top out at 13.65 mm, 11 coins start at 14.85 mm, and 9 or 10 coins can never land on a 0.20 mm mark at 14. Exactly one choice survives both filters, which is the mark of a well-posed count.
💡Key takeaway

Every coin is a dime plus a few 0.20 mm shims, so the height must land on a 0.20 mm mark — that plus a thinnest/thickest squeeze leaves only 8 coins (eight quarters make exactly 14 mm).

  • Measure each coin above a dime
  • Trap the count between 8 and 10
  • Keep only counts that fit the 0.20 step
  • Confirm a real stack of 8 exists

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