AMC 8 · 2014 · #2
Grade 3 arithmeticProblem
Paul owes Paula 35 cents and has a pocket full of 5-cent coins, 10-cent coins, and 25-cent coins that he can use to pay her. What is the difference between the largest and the smallest number of coins he can use to pay her?
Pick an answer.
AMC 8 2014 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: Paul has unlimited $5$-cent, $10$-cent, and $25$-cent coins and needs to pay Paula exactly $35$ cents. Among all coin combinations that total $35$ cents, find the difference between the largest possible number of coins used and the smallest possible number.
Givens: Target amount: $35$ cents; Available coin values: $5$, $10$, $25$ cents; Unlimited supply of each coin ("a pocket full"); Answer choices: (A) $1$, (B) $2$, (C) $3$, (D) $4$, (E) $5$
Unknowns: Largest number of coins that sum to $35$ cents; Smallest number of coins that sum to $35$ cents; Their difference
Understand
Restated: Paul has unlimited $5$-cent, $10$-cent, and $25$-cent coins and needs to pay Paula exactly $35$ cents. Among all coin combinations that total $35$ cents, find the difference between the largest possible number of coins used and the smallest possible number.
Givens: Target amount: $35$ cents; Available coin values: $5$, $10$, $25$ cents; Unlimited supply of each coin ("a pocket full"); Answer choices: (A) $1$, (B) $2$, (C) $3$, (D) $4$, (E) $5$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #17 Pursue Parity
"Largest minus smallest" is really two separate optimization problems wrapped in one subtraction, so Tool #7 (Identify Subproblems) splits it cleanly: maximize the coin count, then minimize it, then subtract. For each sub-question the move is opposite — to maximize the count, use the smallest-value coin as often as possible; to minimize, use the largest-value coin first (a greedy step). Tool #17 (Pursue Parity / divisibility) confirms both extremes are actually reachable: $35$ is a multiple of $5$, so an all-nickel payment works, and $35 = 25 + 10$ gives a clean two-coin payment.
Execute — Answer: E
2.OA.A.1 Step 1 - Subproblem A — minimize the coin count.
- Use the largest coin first.
- One $25$-cent coin leaves $35 - 25 = 10$ cents, which one $10$-cent coin covers exactly.
- That is $2$ coins total, and no single coin equals $35$ cents, so $2$ is the minimum.
💡 Adding two values to hit a target total is a Grade 2 word-problem skill — once you spot $25 + 10 = 35$ you are done.
3.OA.B.6 Step 2 - Subproblem B — maximize the coin count.
- Use the smallest coin (the $5$-cent nickel) as much as possible.
- Since $35$ is divisible by $5$, we can pay entirely in nickels: $35 \div 5 = 7$ coins.
- Using any $10$- or $25$-cent coin would only replace several nickels, lowering the count, so $7$ is the maximum.
💡 Asking "how many $5$s fit into $35$?" is a Grade 3 division-as-unknown-factor question — the divisibility of $35$ by $5$ is exactly the parity / multiple check from Tool #17.
2.OA.A.1 Step 3 Combine the two subproblem answers with subtraction to get the requested difference.
💡 Subtracting the smaller count from the larger is the final "how many more?" step, a Grade 2 subtraction within $100$.
2.OA.A.1 Subproblem A — minimize the coin count. Use the largest coin first. One $25$-cen 3.OA.B.6 Subproblem B — maximize the coin count. Use the smallest coin (the $5$-cent nick 2.OA.A.1 Combine the two subproblem answers with subtraction to get the requested differe Review
Reasonableness: Sanity check both endpoints. Minimum: $25 + 10 = 35$ uses $2$ coins; no other pair of allowed coins sums to $35$ (e.g. $25 + 5 = 30$, $10 + 10 = 20$), so we cannot do it in $1$ coin and $2$ is truly minimal. Maximum: $7 \times 5 = 35$ uses $7$ nickels; any swap of a nickel for a bigger coin (e.g. replace two nickels with one dime) strictly reduces the count, so $7$ is truly maximal. Difference $7 - 2 = 5$ matches choice (E).
Alternative: Tool #6 (Guess and Check) on small coin counts: $1$ coin can only make $5$, $10$, or $25$ — none equal $35$. $2$ coins: $25 + 10 = 35$ works → min $= 2$. For the max, list multiples of $5$ up to $35$ paid only in nickels: $7$ nickels gives $35$, and any dime or quarter would force fewer than $7$ coins, so max $= 7$. Difference $= 5$.
CCSS standards used (min grade 3)
2.OA.A.1Use addition and subtraction within $100$ to solve word problems (Adding $25 + 10 = 35$ to find the minimum coin count, and subtracting $7 - 2 = 5$ to get the final difference.)3.OA.B.6Understand division as an unknown-factor problem (Computing $35 \div 5 = 7$ to count how many $5$-cent coins fit into $35$ cents, giving the maximum coin count.)
⭐ This AMC 8 problem only needs Grade 3 division ($35 \div 5$) plus a tiny bit of Grade 2 adding and subtracting — split it into "fewest coins" and "most coins" and the rest is one quick subtraction.
⭐ This AMC 8 problem only needs Grade 3 division ($35 \div 5$) plus a tiny bit of Grade 2 adding and subtracting — split it into "fewest coins" and "most coins" and the rest is one quick subtraction.
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