AMC 8 · 2010 · #7
Easy mode Grade 3Problem
Freddie has only four kinds of coins: pennies (1 cent), nickels (5 cents), dimes (10 cents), and quarters (25 cents).
He wants to be able to pay any whole-cent amount less than a dollar — like 1 cent, 37 cents, or 99 cents — using the coins he is carrying.
What is the smallest number of coins he needs to carry so this is possible?
Pick an answer.
AMC 8 2010 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.
Toolkit + CCSS Solution
Understand
Restated: Freddie keeps a fixed collection of U.S. coins — only pennies ($1$¢), nickels ($5$¢), dimes ($10$¢), and quarters ($25$¢). He wants the smallest such collection so that for any whole-cent amount from $1$¢ to $99$¢, he can hand over exact change from his collection. What is the smallest total number of coins?
Givens: Allowed coins: penny ($1$¢), nickel ($5$¢), dime ($10$¢), quarter ($25$¢); He must be able to pay every whole-cent amount from $1$¢ up to $99$¢ (less than $\$1$); Answer choices: (A) $6$, (B) $10$, (C) $15$, (D) $25$, (E) $99$
Unknowns: The minimum number of coins Freddie needs in his collection
Understand
Restated: Freddie keeps a fixed collection of U.S. coins — only pennies ($1$¢), nickels ($5$¢), dimes ($10$¢), and quarters ($25$¢). He wants the smallest such collection so that for any whole-cent amount from $1$¢ to $99$¢, he can hand over exact change from his collection. What is the smallest total number of coins?
Givens: Allowed coins: penny ($1$¢), nickel ($5$¢), dime ($10$¢), quarter ($25$¢); He must be able to pay every whole-cent amount from $1$¢ up to $99$¢ (less than $\$1$); Answer choices: (A) $6$, (B) $10$, (C) $15$, (D) $25$, (E) $99$
Plan
Primary tool: #9 Solve an Easier Related Problem
Secondary: #7 Identify Subproblems, #2 Make a Systematic List, #6 Guess and Check
The full range $1$ to $99$ is intimidating, so use Tool #9: solve the easier problem of paying every amount from $1$¢ to $24$¢ first — that locks in the small coins. Tool #2 (Systematic List) is the workhorse inside that easier case — list the multiples of $5$ up to $20$ and check which $\{\text{nickel}, \text{dime}\}$ combos hit each one. Then Tool #7 (Subproblems) handles the upper amounts: once we can cover $0$–$24$¢, adding one quarter at a time extends the range by $25$¢, so $3$ quarters reach all the way to $99$¢. Tool #6 (Guess and Check) is the fast sanity test against the answer choices — $(A)\,6$ is clearly too few, $(C)\,15$ and $(D)\,25$ are wasteful, so the smallest plausible number is around $10$.
Execute — Answer: B
2.MD.C.8 Step 1 - 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.
2.MD.C.8 Step 2 - 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).
2.NBT.B.5 Step 3 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.
3.OA.D.8 Step 4 - 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 $25$¢.
2.NBT.B.5 Step 5 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.
2.MD.C.8 Easier problem first: what is the smallest set of pennies, nickels, and dimes th 2.MD.C.8 Now fill in the $5$, $10$, $15$, $20$ multiples-of-$5$ amounts. One nickel gives 2.NBT.B.5 Subtotal so far: $4$ pennies $+ 1$ nickel $+ 2$ dimes $= 7$ coins, and they can 3.OA.D.8 Extend with quarters. Adding one quarter shifts the covered range by $+25$¢, so 2.NBT.B.5 Total the collection and check the answer choices. Review
Reasonableness: Spot-check a few amounts. $7$¢ $= 1$ nickel $+ 2$ pennies. $38$¢ $= 1$ quarter $+ 1$ dime $+ 3$ pennies. $99$¢ $= 3$ quarters $+ 2$ dimes $+ 4$ pennies — uses every coin in the collection, so we cannot drop any of them. Trying to remove one penny breaks $4$¢; removing the nickel breaks $5$¢ (we have no other way to make $5$ without using $5$ pennies); removing a dime breaks $20$¢; removing a quarter breaks $75$¢. So $10$ is tight, and the answer is (B).
Alternative: Tool #6 (Guess and Check) against the choices: (A) $6$ coins is impossible because we already need $4$ pennies just to handle $4$¢. (C) $15$ and (D) $25$ and (E) $99$ can each be hit by safer-but-wasteful schemes (e.g., $99$ pennies is choice (E)), but none of them is the minimum once we know $10$ works. So (B) is the smallest workable count.
CCSS standards used (min grade 3)
2.MD.C.8Solve word problems involving dollar bills, quarters, dimes, nickels, and pennies (Knowing the U.S. coin values ($1$¢, $5$¢, $10$¢, $25$¢) and building amounts from $1$¢ to $24$¢ out of pennies, nickels, and dimes.)2.NBT.B.5Fluently add and subtract within $100$ (Adding coin values to confirm the collection covers each amount (e.g., $25 + 10 + 3 = 38$¢, total coin count $4+1+2+3=10$).)3.OA.D.8Solve two-step word problems using the four operations (Combining the two subproblems — (i) the smallest small-coin set that covers $0$–$24$¢ and (ii) how many quarters extend the reach to $99$¢ — into one count.)
⭐ This AMC 8 problem only needs Grade 3 reasoning: solve the easier $0$–$24$¢ case with pennies, nickels, and dimes, then let quarters do the rest.
⭐ This AMC 8 problem only needs Grade 3 reasoning: solve the easier $0$–$24$¢ case with pennies, nickels, and dimes, then let quarters do the rest.
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