AMC 8 · 2014 · #2

Grade 3 arithmetic

Archetype Small-Domain Constrained Search

multiplesdivisibility-rulesmental-arithmetic systematic-enumerationidentify-subproblems ↑ Prerequisites: multi-digit-arithmeticmultiples

📏 Short solution 💡 2 insights

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Problem

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?

$\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad \textbf{(E) }5$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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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

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

#7 Identify Subproblems 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.

$$25 + 10 = 35 \;\Rightarrow\; \text{min coins} = 2$$

💡 Adding two values to hit a target total is a Grade 2 word-problem skill — once you spot $25 + 10 = 35$ you are done.

#17 Pursue Parity 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.

$$35 \div 5 = 7 \;\Rightarrow\; \text{max coins} = 7$$

💡 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.

#7 Identify Subproblems 2.OA.A.1 Step 3

Combine the two subproblem answers with subtraction to get the requested difference.

$$7 - 2 = 5 \;\Rightarrow\; \textbf{(E)}$$

💡 Subtracting the smaller count from the larger is the final "how many more?" step, a Grade 2 subtraction within $100$.

[1] #7 2.OA.A.1 Subproblem A — minimize the coin count. Use the largest coin first. One $25$-cen

[2] #17 3.OA.B.6 Subproblem B — maximize the coin count. Use the smallest coin (the $5$-cent nick

[3] #7 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.1 Use 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.6 Understand 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.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: E

Review

CCSS standards used (min grade 3)

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