AMC 8 · 2010 · #7

• Easier problem first: what is the smallest set of pennies, nickels, and dimes that can pay every amount from $1$¢ to $24$¢?
• The ones digit ($0$–$4$) must come from pennies, so we need $4$ pennies.
• The remaining $5$, $10$, $15$, $20$ jumps must come from nickels and dimes.

💡 If you only had $3$ pennies, you could never make $4$¢ — pennies are the only coin worth less than a nickel.

• Now fill in the $5$, $10$, $15$, $20$ multiples-of-$5$ amounts.
• One nickel gives $5$¢.
• One dime gives $10$¢.
• Nickel $+$ dime gives $15$¢.
• Two dimes give $20$¢.
• So $1$ nickel and $2$ dimes are enough to hit every multiple of $5$ from $5$¢ to $20$¢.
• (Only $1$ dime would leave $20$¢ uncoverable — you would need two nickels instead, which is more coins.)

💡 Listing the multiples of $5$ up to $20$ and checking which $\{N, D\}$ combos hit each one is a tiny Tool #2 (systematic list).

Subtotal so far: $4$ pennies $+ 1$ nickel $+ 2$ dimes $= 7$ coins, and they can build every amount from $0$¢ to $24$¢ (penny digit picks $0$–$4$, nickel/dime combo picks $0$, $5$, $10$, $15$, $20$).

💡 Splitting the amount into (ones-digit) + (multiple of $5$) is a clean Tool #7 subproblem split.

• Extend with quarters.
• Adding one quarter shifts the covered range by $+25$¢, so $1$ quarter covers $25$–$49$¢, $2$ quarters cover $50$–$74$¢, $3$ quarters cover $75$–$99$¢.
• Three quarters are exactly enough to reach $99$¢ (and we never need to make a full dollar).

💡 Each new quarter is a self-contained subproblem: it pushes the upper reach of the collection up by $25$¢.

Total the collection and check the answer choices.

💡 $(A)\,6$ is too few (you already need $4$ pennies and $3$ quarters), and the bigger choices are clearly wasteful — $(B)\,10$ is the only one that matches our build.