AMC 8 · 2020 · #8

Grade 5 arithmetic

Archetype Arithmetic Expression Decomposition

multi-digit-arithmeticmental-arithmetic identify-subproblemsbound-inequality-then-enumerate ↑ Prerequisites: multi-digit-arithmetic

📏 Medium solution 💡 3 insights

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Problem

Ricardo has $2020$ coins, some of which are pennies ( $1$ -cent coins) and the rest of which are nickels ( $5$ -cent coins). He has at least one penny and at least one nickel. What is the difference in cents between the greatest possible and least amounts of money that Ricardo can have?

$\textbf{(A) }\text{806} \qquad \textbf{(B) }\text{8068} \qquad \textbf{(C) }\text{8072} \qquad \textbf{(D) }\text{8076}\qquad \textbf{(E) }\text{8082}$

Pick an answer.

(A)

$text{806}$

(B)

$text{8068}$

(C)

$text{8072}$

(D)

$text{8076}$

(E)

$text{8082}$

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Toolkit + CCSS Solution

Understand

Restated: Ricardo has exactly $2020$ coins made up of pennies ($1$ cent each) and nickels ($5$ cents each), with at least one of each kind. Find the difference, in cents, between the greatest possible total value and the least possible total value of his coin collection.

Givens: Total number of coins is $2020$; Pennies are worth $1$ cent each; nickels are worth $5$ cents each; At least one penny AND at least one nickel; Answer choices: (A) $806$, (B) $8068$, (C) $8072$, (D) $8076$, (E) $8082$ (cents)

Unknowns: The greatest possible total value (in cents); The least possible total value (in cents); The difference between the greatest and the least

Understand

Plan

Primary tool: #7 Identify Subproblems

Secondary: #9 Solve an Easier Related Problem, #6 Guess and Check

The question hides three smaller jobs inside one sentence: (a) what is the largest total value possible, (b) what is the smallest total value possible, and (c) what is their difference. Tool #7 (Identify Subproblems) splits the problem into those three clean steps. To see which mix of coins makes the value largest or smallest, Tool #9 (Easier Related Problem) is handy — try just $5$ coins with at-least-one-of-each, and the pattern "put all but one into the high-value coin (or low-value coin)" becomes obvious. Tool #6 (Guess and Check) lets us confirm the extreme by comparing one trial swap (e.g., $2$ pennies and $2018$ nickels) against the candidate extreme.

Execute — Answer: C

#9 Solve an Easier Related Problem 5.NBT.B.5 Step 1

Use a smaller version first: suppose Ricardo had only $5$ coins with at least one of each.
To maximize value, put $1$ penny and $4$ nickels: value $= 1 + 20 = 21$ cents.
Any swap of a nickel for an extra penny loses $4$ cents, so this really is the max.
The pattern generalizes: with $2020$ coins, the maximum uses exactly $1$ penny and the rest nickels.

$$\text{Max} = 1 \times 1 + 2019 \times 5 = 1 + 10095 = 10096 \text{ cents}$$

💡 Multiplying $2019 \times 5$ is a Grade 5 multi-digit multiplication.

#7 Identify Subproblems 4.OA.A.3 Step 2

By the same logic, to minimize value, push all the weight to the cheaper coin: $1$ nickel and the remaining $2019$ pennies.
Swapping a penny for a nickel would add $4$ cents, so this is genuinely the minimum.

$$\text{Min} = 2019 \times 1 + 1 \times 5 = 2019 + 5 = 2024 \text{ cents}$$

💡 This is the second subproblem — a multi-step word problem using four operations, a Grade 4 skill.

#6 Guess and Check 4.NBT.B.4 Step 3

Confirm the extremes with Tool #6 (Guess and Check) using one neighbouring split: $2$ pennies and $2018$ nickels gives $2 + 10090 = 10092$ cents — that is $4$ cents less than $10096$, so moving a nickel into a penny costs $4$ cents every time.
The maximum and minimum found above are correct.

$$10096 - 10092 = 4 \text{ cents per swap}$$

💡 Comparing two big totals with a quick subtraction is Grade 4 multi-digit arithmetic.

#7 Identify Subproblems 4.NBT.B.4 Step 4

Subtract the minimum from the maximum to answer the actual question.

$$10096 - 2024 = 8072 \text{ cents}$$

💡 Subtracting two $4$- and $5$-digit numbers is Grade 4 multi-digit subtraction.

#7 Identify Subproblems 4.OA.A.3 Step 5

Match the numerical answer to the multiple-choice list.
$8072$ appears as choice (C).

$$\text{Difference} = 8072 \;\Rightarrow\; \textbf{(C)}$$

💡 Reading and selecting the matching answer is the final step of any multi-step word problem (Grade 4).

[1] #9 5.NBT.B.5 Use a smaller version first: suppose Ricardo had only $5$ coins with at least on

[2] #7 4.OA.A.3 By the same logic, to minimize value, push all the weight to the cheaper coin: $

[3] #6 4.NBT.B.4 Confirm the extremes with Tool #6 (Guess and Check) using one neighbouring split

[4] #7 4.NBT.B.4 Subtract the minimum from the maximum to answer the actual question.

[5] #7 4.OA.A.3 Match the numerical answer to the multiple-choice list. $8072$ appears as choice

Review

Reasonableness: A neat shortcut confirms the answer. Going from the min split ($2019$ pennies, $1$ nickel) to the max split ($1$ penny, $2019$ nickels) flips the role of $2018$ coins from penny to nickel; each flip adds $5 - 1 = 4$ cents. So the difference must be $2018 \times 4 = 8072$ cents, matching the direct subtraction. Both the size ($\sim 8000$) and the parity (multiple of $4$) make sense, and $8072$ is one of the given choices, so the answer is consistent.

Alternative: Tool #13 (Convert to Algebra) also works: let $p$ be the number of pennies, so nickels $= 2020 - p$ and total value $V(p) = p + 5(2020 - p) = 10100 - 4p$. Because $V$ decreases by $4$ for each extra penny, $V$ is largest when $p$ is smallest ($p = 1$, $V = 10096$) and smallest when $p$ is largest ($p = 2019$, $V = 2024$). The difference is $4 \times (2019 - 1) = 8072$. This is heavier machinery than the subproblems approach, so prefer Tool #7 for a younger learner.

CCSS standards used (min grade 5)

4.OA.A.3 Solve multi-step word problems using four operations with whole numbers (Treating the question as a multi-step word problem — find max, find min, take their difference, match to a multiple-choice option.)
4.NBT.B.4 Fluently add and subtract multi-digit whole numbers (Computing $10096 - 10092 = 4$ to confirm the per-swap penalty and $10096 - 2024 = 8072$ for the final difference.)
5.NBT.B.5 Fluently multiply multi-digit whole numbers (Multiplying $2019 \times 5 = 10095$ when calculating the maximum total value.)

⭐ This AMC 8 problem only needs Grade 5 multi-digit multiplication and a little Grade 4 word-problem thinking you already know!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 5)

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