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

In the United States, coins have the following thicknesses: penny, $1.55$ mm; nickel, $1.95$ mm; dime, $1.35$ mm; quarter, $1.75$ mm. If a stack of these coins is exactly $14$ mm high, how many coins are in the stack?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2004 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.

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Toolkit + CCSS Solution

Understand

Restated: Coins come in four thicknesses: penny $1.55$ mm, nickel $1.95$ mm, dime $1.35$ mm, quarter $1.75$ mm. Some coins are stacked and the pile is exactly $14$ mm tall. Find how many coins are in the stack.

Givens: Penny is $1.55$ mm thick.; Nickel is $1.95$ mm thick.; Dime is $1.35$ mm thick.; Quarter is $1.75$ mm thick.; The stack is exactly $14$ mm high.; Answer choices: (A) $7$, (B) $8$, (C) $9$, (D) $10$, (E) $11$.

Unknowns: The number of coins in the stack.

Understand

Plan

Primary tool: #3 Eliminate Possibilities

Secondary: #8 Analyze the Units, #14 Extreme Principle

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.

Execute — Answer: B

#8 Analyze the Units 5.NBT.B.7 Step 1

Measure each coin above a dime

The dime is the thinnest coin at $1.35$ mm, so compare the others to it.
Penny $-$ dime $= 1.55 - 1.35 = 0.20$.
Quarter $-$ dime $= 1.75 - 1.35 = 0.40$.
Nickel $-$ dime $= 1.95 - 1.35 = 0.60$.
Every extra ($0$, $0.20$, $0.40$, $0.60$) is a whole number of $0.20$ mm steps.
So a stack of $n$ coins is $n$ dimes ($1.35n$) plus some whole number of $0.20$ mm steps: its height is $1.35n + 0.20 \times (\text{a whole number})$.

$$0.20,\ 0.40,\ 0.60 = 1\times 0.20,\ 2\times 0.20,\ 3\times 0.20 \ \Rightarrow\ \text{height} = 1.35n + 0.20\times(\text{whole number})$$

💡 Every coin is just a dime with a few $0.20$ mm shims on top, so only the shims change the height.

#14 Extreme Principle 6.NS.B.3 Step 2

Trap the count between 8 and 10

The thinnest stack of $n$ coins is all dimes ($1.35n$) and the thickest is all nickels ($1.95n$); the real stack sits between them.
To reach $14$, even all nickels must be tall enough: $1.95 \times 7 = 13.65 < 14$, so $7$ coins can never reach $14$ and $n \ge 8$.
And even all dimes must not overshoot: $1.35 \times 11 = 14.85 > 14$, so $11$ coins always pass $14$ and $n \le 10$.
That eliminates (A) $7$ and (E) $11$, leaving $n = 8$, $9$, or $10$.

$$1.95\times 7 = 13.65 < 14 \le 1.95n \Rightarrow n\ge 8;\quad 1.35n \le 14 < 14.85 = 1.35\times 11 \Rightarrow n\le 10$$

💡 Too few coins can't reach $14$ mm even at their thickest, and too many overshoot even at their thinnest.

#3 Eliminate Possibilities 4.OA.B.4 Step 3

Keep only counts that fit the 0.20 step

From Step 1 the height is $1.35n + 0.20\times(\text{whole number})$, so $14 - 1.35n$ must be a non-negative multiple of $0.20$.
Test the survivors.
$n=8$: $14 - 1.35\times 8 = 14 - 10.80 = 3.20 = 0.20\times 16$ — a whole number of steps, good.
$n=9$: $14 - 12.15 = 1.85$, and $1.85 \div 0.20 = 9.25$ — not whole, out.
$n=10$: $14 - 13.50 = 0.50$, and $0.50 \div 0.20 = 2.5$ — not whole, out.
Only $n=8$ passes.

$$n=8:\ 3.20=0.20\times 16\ \checkmark\qquad n=9:\ 1.85=0.20\times 9.25\ \times\qquad n=10:\ 0.50=0.20\times 2.5\ \times$$

💡 Heights only land on $0.20$ mm marks, so a leftover that isn't a whole number of steps can't be built.

#3 Eliminate Possibilities 5.NBT.B.7 Step 4

Confirm a real stack of 8 exists

A count only counts if some real stack actually reaches $14$ mm.
For $n=8$ the shims must total $0.20\times 16 = 3.20$ mm above eight dimes.
The simplest choice: make every coin a quarter, which is $0.40$ above a dime.
Eight quarters give shims $8\times 0.40 = 3.20$ mm, and directly $8 \times 1.75 = 14.00$ mm.
So $8$ coins work exactly, and it is the only surviving count.
The answer is (B).

$$8\times 1.75 = 14.00\text{ mm}\ \Rightarrow\ n = 8\ \textbf{(B)}$$

💡 Eight quarters stack to exactly $14$ mm, so the count of $8$ isn't just allowed — it really happens.

[1] #8 5.NBT.B.7 The dime is the thinnest coin at $1.35$ mm, so compare the others to it. Penny $

[2] #14 6.NS.B.3 The thinnest stack of $n$ coins is all dimes ($1.35n$) and the thickest is all n

[3] #3 4.OA.B.4 From Step 1 the height is $1.35n + 0.20\times(\text{whole number})$, so $14 - 1.

[4] #3 5.NBT.B.7 A count only counts if some real stack actually reaches $14$ mm. For $n=8$ the s

Review

Reasonableness: Eight quarters give $8\times 1.75 = 14$ 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.

Alternative: Scale away the decimals. In hundredths of a millimeter the thicknesses are $155, 195, 135, 175$ and the target is $1400$; dividing by $5$ gives $31, 39, 27, 35$ with target $280$. Each of $31, 39, 27, 35$ leaves remainder $3$ when divided by $4$, so $n$ coins total $3n \pmod 4$. Since $280 = 4\times 70$ is a multiple of $4$, we need $3n \equiv 0 \pmod 4$, i.e. $n$ is a multiple of $4$. Combined with $8 \le n \le 10$, the only multiple of $4$ is $n = 8$.

CCSS standards used (min grade 6)

5.NBT.B.7 Add, subtract, multiply, and divide decimals to hundredths (Subtracting to find each coin's $0.20$ mm steps above a dime and checking $8\times 1.75 = 14$.)
6.NS.B.3 Fluently add, subtract, multiply, and divide multi-digit decimals (Multiplying thicknesses like $1.95\times 7$ and $1.35\times 11$ to trap the coin count between $8$ and $10$.)
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite (Testing whether $14 - 1.35n$ is a whole-number multiple of the $0.20$ mm step to eliminate $n=9$ and $n=10$.)

⭐ 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).

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

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

Review

CCSS standards used (min grade 6)

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