2004 AMC 10B Problem 13

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 $\boxed{\textbf{(B)}\ 8}$.

When several allowed step sizes share a residue modulo some number, the total is determined modulo that number by the count alone. That converts "which coins?" into "how many coins?", which is what the problem asks. Pair the congruence with crude upper and lower bounds and the count is pinned down; the explicit example (eight quarters) confirms the answer is attainable.

Notice that $4$ quarters stack to exactly $7$ mm, so $8$ quarters make $14$ mm. Since the problem promises a unique count, $8$ must be it. This is fast but relies on the problem's guarantee; the mod-$20$ argument is what actually excludes $9$ and $10$.

• Bounding alone (8, 9, or 10 coins all seem possible by height) and guessing, without the mod-20 argument that eliminates 9 and 10.
• Dividing $14$ by an "average" thickness of about $1.65$ mm and rounding to $8$ or $9$ without justification.