2004 AMC 10B Problem 13

P11-15~4 minPrint

Modular Arithmetic Diophantine Equations Bound the quantity above/below or estimate to pin it down Use an invariant, parity, or coloring argument

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?

Show solution without answering

Key insight

In hundredths of a millimeter, every coin is 15 more than a multiple of 20. So the coin count must be a multiple of 4: 8.

Solution

Decimals are awkward, so measure in hundredths of a millimeter. The thicknesses become $155$ (penny), $195$ (nickel), $135$ (dime), $175$ (quarter), and the stack is $1400$ .

Step 1: how many coins could fit by size?

The thinnest coin is $135$ . Ten dimes are $1350$ and eleven are $1485$ , so more than $10$ coins is always too tall.
The thickest coin is $195$ . Seven nickels are $1365$ , still short of $1400$ , so fewer than $8$ coins is always too short.
So the stack has $8$ , $9$ or $10$ coins.

Step 2: find a pattern in the thicknesses.

4. Divide each thickness by $20$ and look at the remainder:
$155 = 7 \cdot 20 + 15, \quad 195 = 9 \cdot 20 + 15, \quad 135 = 6 \cdot 20 + 15, \quad 175 = 8 \cdot 20 + 15.$
Every coin is "some twenties plus $15$ ."
5. So a stack of $n$ coins is some twenties plus $15n$ . For the stack to be $1400$ , which is a whole number of twenties ( $70 \times 20$ ), the leftover $15n$ must also be a whole number of twenties.
6. Check the three candidates: $15 \times 8 = 120 = 6 \times 20$ works. $15 \times 9 = 135$ is not a multiple of $20$ . $15 \times 10 = 150$ is not a multiple of $20$ either.
7. So the stack must have $8$ coins.

Step 3: make sure it can really happen.

Eight quarters: $8 \times 175 = 1400$ . Exactly $14$ mm.

The answer is $\boxed{\textbf{(B)}\ 8}$ .

Why this works

When every piece has the same remainder, the total's remainder depends only on how many pieces there are, not which ones. That turns "which coins?" into "how many coins?", which is just what the question asks. Rough size limits narrow the choices to a few, and the remainder check picks the one that works.

Alternative approach

Look for an easy stack. Four quarters are $4 \times 1.75 = 7$ mm, so eight quarters are $14$ mm. That gives $8$ quickly, and the problem suggests only one count is possible. The remainder argument above is what shows that $9$ or $10$ coins really cannot work.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Measure in hundredths of a millimeter to avoid decimals: the thicknesses are $155, 195, 135, 175$ and the stack is $1400$ .

Each thickness leaves remainder $15$ when divided by $20$ : $155 = 7 \cdot 20 + 15$ , $195 = 9 \cdot 20 + 15$ , $135 = 6 \cdot 20 + 15$ , $175 = 8 \cdot 20 + 15$ . So no matter which coins are used, a stack of $n$ coins has height $\equiv 15n \pmod{20}$ . The target $1400$ is a multiple of $20$ , so we need
$15n \equiv 0 \pmod{20} \quad\Longleftrightarrow\quad 3n \equiv 0 \pmod 4 \quad\Longleftrightarrow\quad 4 \mid n.$

Next, bound $n$ . The thinnest coin is $135$ and the thickest is $195$ , so $135n \le 1400 \le 195n$ , giving $7.18 \le n \le 10.37$ , i.e. $n \in \{8, 9, 10\}$ . The only multiple of $4$ there is $n = 8$ .

Such a stack exists: eight quarters measure $8 \cdot 175 = 1400$ .

The answer is $\boxed{\textbf{(B)}\ 8}$ .

Review in the AoPS books

Introduction to Number Theory, Chapter 12: Introduction to Modular Arithmetic
Introduction to Number Theory, Chapter 7: Algebra With Integers
Introduction to Number Theory, Chapter 10: Units Digits
Introduction to Algebra, Chapter 11: Special Factorizations (§11.5)