AMC 10 · 2023 · #11

Grade 6 arithmetic

linear-diophantinesystematic-enumerationlinear-equations-two-varbound-inequality-then-enumerate systematic-enumerationcaseworkbound-inequality-then-enumerate ↑ Prerequisites: linear-equations-two-varsystematic-enumeration

📏 Medium solution 💡 3 insights

Problem

Suzanne went to the bank and withdrew $800$. The teller gave her this amount using$ 20 $bills,$ $50$ bills, and $$100$ bills, with at least one of each denomination. How many different collections of bills could Suzanne have received?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2023 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.

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Toolkit + CCSS Solution

Understand

Restated: Suzanne withdraws $$800$ from the bank. The teller hands it over using only $$20$, $$50$, and $$100$ bills, with at least one bill of every kind. Count the number of different stacks (combinations of bill counts) she could receive.

Givens: Total cash handed over: $$800$; Allowed denominations: $$20$, $$50$, $$100$ bills; At least one of each denomination is included; Answer choices: (A) $45$, (B) $21$, (C) $36$, (D) $28$, (E) $32$

Unknowns: Number of ordered triples $(x, y, z)$ of positive integers with $20x + 50y + 100z = 800$

Understand

Plan

Primary tool: #2 Make a Systematic List

Secondary: #9 Solve an Easier Related Problem, #5 Look for a Pattern, #3 Eliminate Possibilities

Tool #2 (Systematic List) is the natural fit because the question literally asks "how many collections" — that triggers an organized count. Tool #9 (Easier Problem) helps: dividing the equation by $10$ shrinks $20x + 50y + 100z = 800$ to $2x + 5y + 10z = 80$, the same problem with smaller numbers. To avoid listing all 21 by hand, Tool #5 (Pattern) catches that the count of $(x, y)$ pairs for each fixed $z$ is an arithmetic sequence — its sum is the answer.

Execute — Answer: B

#9 Solve an Easier Related Problem 6.EE.A.3 Step 1

Shrink the equation first.
Divide $20x + 50y + 100z = 800$ by $10$ to get a friendlier equation with the exact same solutions.

$$2x + 5y + 10z = 80, \quad x, y, z \ge 1$$

💡 Dividing every term by $10$ keeps the equation balanced — a Grade 6 "equivalent expression" move that makes the numbers easier.

#2 Make a Systematic List 6.EE.B.6 Step 2

Order the count by the largest bill first.
Fix $z$ (the number of $$100$s) and count how many $(x, y)$ work. For each $z$, the equation $2x + 5y = 80 - 10z$ has $x \ge 1$ and $y \ge 1$, and $2x$ must equal $80 - 10z - 5y$, so $5y$ must have the same parity as $80 - 10z$, which is even — meaning $y$ must be even.

Fix $z$. Need $2x + 5y = 80 - 10z$ with $x \ge 1$, $y \ge 1$, $y$ even.

💡 Pick an outer variable to fix (here $z$) so the inner equation becomes a two-variable problem you can count — Grade 6 use of variables.

#5 Look for a Pattern 6.EE.B.6 Step 3

For each fixed $z$, write $y = 2b$ ($b \ge 1$) and check what $x$ must be.
The equation becomes $2x + 10b = 80 - 10z$, i.e.
$x = 40 - 5z - 5b = 5(8 - z - b)$.
So $x$ must be a positive multiple of $5$: $x \ge 5$, and that forces $8 - z - b \ge 1$, i.e.
$b \le 7 - z$.
Since $b \ge 1$, the valid $b$ count is $7 - z$ — provided $z \le 6$.

Number of pairs for fixed $z$: $\max(7 - z, 0)$.

💡 Each value of $z$ gives a clean linear count of valid $b$'s; recognizing the arithmetic-decay pattern is Tool #5.

#2 Make a Systematic List 6.EE.B.7 Step 4

List the counts for $z = 1, 2, 3, 4, 5, 6$ (and $z = 7$ gives $0$).
Add them.

$$6 + 5 + 4 + 3 + 2 + 1 = 21$$

💡 Adding the per-$z$ counts is the standard "sum the cases" closeout for systematic counting.

#3 Eliminate Possibilities 6.EE.B.5 Step 5

Match to the choices: $21$ is choice (B).
The other choices ($45$, $36$, $28$, $32$) do not match any natural undercount or overcount of the cases — they are distractors.

$$\text{Answer} = 21 \;\Rightarrow\; \textbf{(B)}$$

💡 Comparing the computed total to the five choices is the standard multiple-choice finish.

[1] #9 6.EE.A.3 Shrink the equation first. Divide $20x + 50y + 100z = 800$ by $10$ to get a frie

[2] #2 6.EE.B.6 Order the count by the largest bill first. Fix $z$ (the number of $$100$s) and c

[3] #5 6.EE.B.6 For each fixed $z$, write $y = 2b$ ($b \ge 1$) and check what $x$ must be. The e

[4] #2 6.EE.B.7 List the counts for $z = 1, 2, 3, 4, 5, 6$ (and $z = 7$ gives $0$). Add them.

[5] #3 6.EE.B.5 Match to the choices: $21$ is choice (B). The other choices ($45$, $36$, $28$, $

Review

Reasonableness: Sanity-check the boundary cases. When $z = 1$ (one $$100$), the remaining $$700$ must come from $$20$s and $$50$s with at least one of each: $20x + 50y = 700$ with $x, y \ge 1$. Listing: $y \in \{2, 4, 6, 8, 10, 12\}$ (must be even, and $50y \le 680$), giving $6$ collections — matches the $7 - 1 = 6$ formula. Spot-check $z = 6$ (six $$100$s, $$200$ left): $20x + 50y = 200$, $y \in \{2\}$, $x = 5$ — exactly $1$ collection, matches $7 - 6 = 1$. Total $21$ is small enough to feel right for a "three denominations, $$800$ total" setup.

Alternative: Tool #13 (Convert to Algebra) gives the classic stars-and-bars route: substitute $x = 5a$ (forced because $80 - 5y - 10z$ must be a multiple of $5$) and $y = 2b$ (forced because $2x + 10z$ is even, so $5y$ must be even). The simplified equation $a + b + z = 8$ with all $\ge 1$ has $\binom{7}{2} = 21$ positive-integer solutions. Same answer, faster for an algebra-comfortable student.

CCSS standards used (min grade 6)

6.EE.A.3 Apply the properties of operations to generate equivalent expressions (Dividing $20x + 50y + 100z = 800$ by $10$ to get the equivalent equation $2x + 5y + 10z = 80$.)
6.EE.B.6 Use variables to represent numbers and write expressions to solve problems (Fixing $z$ and writing $y = 2b$ to convert the count into a single-variable check on $b$.)
6.EE.B.7 Solve real-world problems by writing and solving equations of the form px = q (Solving $x = 5(8 - z - b)$ for $x$ in terms of $z$ and $b$ to enumerate valid pairs.)
6.EE.B.5 Understand solving an equation or inequality as a process of finding values (Comparing the summed count $21$ against the five answer choices to pick (B).)

⭐ This AMC 10 problem only needs Grade 6 "equivalent equations and variables" you already know — once you divide the dollar amounts by $10$ and fix the number of $$100$ bills, each case turns into a tidy little count, and the cases add up to $21$.