AMC 8 · 2000 · #20

Easy mode Grade 4

Problem

Imagine you have a pile of $9$ U.S. coins. Some are pennies ( $1$ cent each), some are nickels ( $5$ cents each), some are dimes ( $10$ cents each), and some are quarters ( $25$ cents each).

Two things to remember:

You have at least one of every kind of coin.
All $9$ coins together are worth exactly $ $1.02$ (that is, $102$ cents).

How many of the $9$ coins must be dimes?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

View mode:

Toolkit + CCSS Solution

Understand

Restated: You have exactly $9$ coins made of pennies ($1$¢), nickels ($5$¢), dimes ($10$¢), and quarters ($25$¢), with at least one of each kind. The coins add up to $102$¢. How many dimes must you have?

Givens: Total of $9$ coins; Four coin types: penny $1$¢, nickel $5$¢, dime $10$¢, quarter $25$¢; At least one coin of each type; Total value $= 102$¢; Answer choices: (A) $1$, (B) $2$, (C) $3$, (D) $4$, (E) $5$

Unknowns: The number of dimes

Understand

Plan

Primary tool: #3 Eliminate Possibilities

Secondary: #8 Analyze the Units, #2 Make a Systematic List

First peel off one coin of each kind to satisfy the "at least one" rule. That fixes $41$¢ in $4$ coins and leaves $61$¢ to share among $5$ more coins. Tool #8 (Analyze the Units) gives the key restriction: nickels, dimes, and quarters all contribute multiples of $5$, so the number of extra pennies must make the leftover a multiple of $5$. Tool #3 (Eliminate Possibilities) then knocks out every penny count except one. With pennies pinned down, Tool #2 (Make a Systematic List) tries each possible extra-dime count and eliminates the ones that can't be completed by nickels and quarters. No equations needed — just divisibility and short bookkeeping.

Execute — Answer: A

#3 Eliminate Possibilities 4.MD.A.2 Step 1

Use the "one of each" rule to shrink the problem.
Set aside one penny, one nickel, one dime, and one quarter.
That uses $4$ coins and $1 + 5 + 10 + 25 = 41$¢.
What remains: $5$ more coins totaling $102 - 41 = 61$¢, of the same four types.

$$\text{remaining coins} = 9 - 4 = 5, \quad \text{remaining value} = 102 - 41 = 61\text{¢}$$

💡 Spending the "at least one" requirement up front turns a four-variable puzzle into a smaller one with no minimum constraints.

#8 Analyze the Units 4.OA.B.4 Step 2

Use divisibility to fix the extra pennies.
Nickels, dimes, and quarters are all multiples of $5$¢, so whatever they total ends in $0$ or $5$.
The $61$¢ leftover ends in $1$, so the extra pennies must supply that "$1$" — meaning the number of extra pennies, mod $5$, equals $1$.
The only counts in $\{0,1,2,3,4,5\}$ that work are $1$ or $6$.
Since only $5$ coins remain, $6$ extra pennies is impossible.
So there is exactly $1$ extra penny.

$$\text{extra pennies} \equiv 1 \pmod 5, \quad 0 \le \text{extra pennies} \le 5 \;\Rightarrow\; \text{extra pennies} = 1$$

💡 Looking only at the ones digit (the unit of $1$¢) forces the penny count without touching the other coins.

#3 Eliminate Possibilities 4.MD.A.2 Step 3

Update the remaining budget.
After placing $1$ more penny, $5 - 1 = 4$ coins are still to assign, and the value left is $61 - 1 = 60$¢.
These $4$ coins are nickels, dimes, or quarters only.

$$\text{coins left} = 4, \quad \text{value left} = 60\text{¢}$$

💡 With pennies done, every remaining coin is worth a multiple of $5$¢ — the rest is pure trial on a small board.

#2 Make a Systematic List 4.OA.A.3 Step 4

List the cases by number of extra dimes.
Let $d$ be the number of dimes used among the $4$ remaining coins.
The other $4 - d$ coins are nickels or quarters and must total $60 - 10d$ cents.
Try $d = 0, 1, 2, 3, 4$ and check whether nickels (worth $5$) and quarters (worth $25$) can hit the target with the correct count.\n\n• $d = 4$: $0$ coins must total $60 - 40 = 20$¢.
Impossible.\n• $d = 3$: $1$ coin must total $30$¢.
No single coin equals $30$¢.
Impossible.\n• $d = 2$: $2$ coins must total $40$¢.
Options $5+5=10$, $5+25=30$, $25+25=50$.
None hits $40$.
Impossible.\n• $d = 1$: $3$ coins must total $50$¢.
With $q$ quarters: $25q + 5(3-q) = 15 + 20q = 50 \Rightarrow 20q = 35$.
No integer $q$.
Impossible.\n• $d = 0$: $4$ coins must total $60$¢.
With $q$ quarters: $25q + 5(4-q) = 20 + 20q = 60 \Rightarrow q = 2$, so $n = 2$ nickels.
Works.

$$\text{Only valid case: } d = 0,\ q = 2,\ n = 2$$

💡 Five small cases, each settled by one quick check — exactly the situation a systematic list is built for.

#3 Eliminate Possibilities 4.MD.A.2 Step 5

Add the original dime back in.
The final coin bag is: $1 + 1 = 2$ pennies, $1 + 2 = 3$ nickels, $1 + 0 = 1$ dime, $1 + 2 = 3$ quarters.
Check: $2 + 3 + 1 + 3 = 9$ coins, $2(1) + 3(5) + 1(10) + 3(25) = 2 + 15 + 10 + 75 = 102$¢.
The dime count is $1$.

$$\text{dimes} = 1 \;\Rightarrow\; \textbf{(A)}$$

💡 Don't forget to put back the dime that was set aside at the very start — the count of $1$ comes from the original alone.

[1] #3 4.MD.A.2 Use the "one of each" rule to shrink the problem. Set aside one penny, one nicke

[2] #8 4.OA.B.4 Use divisibility to fix the extra pennies. Nickels, dimes, and quarters are all

[3] #3 4.MD.A.2 Update the remaining budget. After placing $1$ more penny, $5 - 1 = 4$ coins are

[4] #2 4.OA.A.3 List the cases by number of extra dimes. Let $d$ be the number of dimes used amo

[5] #3 4.MD.A.2 Add the original dime back in. The final coin bag is: $1 + 1 = 2$ pennies, $1 +

Review

Reasonableness: The final bag $\{2 \text{ pennies}, 3 \text{ nickels}, 1 \text{ dime}, 3 \text{ quarters}\}$ satisfies every condition: $9$ coins, at least one of each type, and $2 + 15 + 10 + 75 = 102$¢. The case-by-case work also showed it is the *only* bag that works (every other dime count led to a contradiction), so the answer must be unique. A unique number of dimes is exactly what the question asks for — the word "must" only makes sense when the count is forced. Answer $1$ matches choice (A).

Alternative: Tool #6 (Guess and Check) on the answer choices: try each value as the dime count and see whether the remaining coins can be filled in. For $d_{\text{total}} = 1$, one dime contributes $10$¢ and uses $1$ coin, leaving $92$¢ in $8$ coins from pennies/nickels/quarters. A quick search finds $2$ pennies, $3$ nickels, $3$ quarters — total checks out. For $d_{\text{total}} = 2, 3, 4, 5$, similar searches fail to satisfy both the coin count and the total at once (the divisibility argument above explains why). Only (A) survives.

CCSS standards used (min grade 4)

4.MD.A.2 Use the four operations to solve word problems involving money (Adding and subtracting cent values for the four coin types — $41$¢ already placed, $61$¢ remaining, then the final check $2 + 15 + 10 + 75 = 102$¢.)
4.OA.B.4 Recognize that a whole number is a multiple of each of its factors; determine multiples and factors (Using the fact that nickels, dimes, and quarters all contribute multiples of $5$¢ to force the number of extra pennies to be $\equiv 1 \pmod 5$.)
4.OA.A.3 Solve multi-step word problems with whole numbers using the four operations (Walking through the five possible dime counts ($d = 0, 1, 2, 3, 4$) and checking each with multiplication and subtraction to find the only feasible case.)

⭐ Hand out one of each coin first to clear the "at least one" rule. The leftover $61$¢ ends in a $1$, so only pennies can supply that odd unit — and with just $5$ coins left, there's room for exactly $1$ extra penny. From there, a short check of dime counts $0$ through $4$ leaves only one working bag, with $1$ dime total — answer (A).

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: A

Review

CCSS standards used (min grade 4)

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