AMC 10 · 2020 · #7

Grade 5 arithmetic

Archetype Small-Domain Constrained Search └ Sub-archetype Factor-Pair Enumeration

multiplesperfect-squaresdivisibility-rulesbound-inequality-then-enumerate easier-related-problembound-inequality-then-enumeratesystematic-enumeration ↑ Prerequisites: perfect-squaresdivisibility-rules

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Problem

How many positive even multiples of $3$ less than $2020$ are perfect squares?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2020 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: Count the positive perfect squares that are less than $2020$ and are also even multiples of $3$.

Givens: The number must be a positive perfect square; The number must be less than $2020$; The number must be even (a multiple of $2$); The number must be a multiple of $3$; Choices: (A) $7$, (B) $8$, (C) $9$, (D) $10$, (E) $12$

Unknowns: The count of such perfect squares

Understand

Restated: Count the positive perfect squares that are less than $2020$ and are also even multiples of $3$.

Plan

Primary tool: #2 Make a Systematic List

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

Tool #9 (Easier) replaces "even multiple of $3$" with the simpler equivalent "multiple of $6$". Tool #2 then lists the candidates $6^2, 12^2, 18^2, \ldots$ in order and stops when they exceed $2020$. Tool #5 reveals these are exactly the squares $(6k)^2 = 36k^2$, so we just need to count how many $k$ work. Tool #3 confirms against the answer choices.

Execute — Answer: A

#9 Solve an Easier Related Problem 4.OA.B.4 Step 1

Recognize that "even multiple of $3$" $=$ "multiple of $2$ AND multiple of $3$" $=$ "multiple of $6$".
So we need perfect squares less than $2020$ that are multiples of $6$.

$$\text{even} \cap \text{multiple of }3 = \text{multiple of }6$$

💡 Simpler phrasing: just multiples of $6$.

#5 Look for a Pattern 4.OA.B.4 Step 2

If $n^2$ is a multiple of $6$, then $n$ itself must be a multiple of $6$.
Reason: $6 = 2 \cdot 3$ with $2$ and $3$ prime, so each prime factor of $6$ that divides $n^2$ must already divide $n$.
So we are counting $n = 6, 12, 18, 24, \ldots$ with $n^2 < 2020$.

$$n = 6k, \quad k \in \{1, 2, 3, \ldots\}, \quad n^2 = 36k^2 < 2020$$

💡 $n^2$ multiple of $6$ $\Rightarrow$ $n$ multiple of $6$.

#2 Make a Systematic List 5.NBT.B.5 Step 3

List the squares in order until one passes $2020$.
With $n = 6k$, compute $36k^2$ for $k = 1, 2, 3, \ldots$

$$k=1: 36,\;\; k=2: 144,\;\; k=3: 324,\;\; k=4: 576,\;\; k=5: 900,\;\; k=6: 1296,\;\; k=7: 1764,\;\; k=8: 2304$$

💡 Generate candidates in order from smallest.

#2 Make a Systematic List 4.NBT.A.2 Step 4

Compare each entry to $2020$.
The first seven ($36, 144, 324, 576, 900, 1296, 1764$) all stay under $2020$.
The eighth ($2304$) overshoots.
So $k = 1$ through $k = 7$ work — that is $7$ values.

$$1764 < 2020 < 2304 \;\Rightarrow\; k \in \{1, 2, \ldots, 7\}$$

💡 Stop the list at the first miss.

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

$7$ matches choice (A).

$$7 \Rightarrow \textbf{(A)}$$

💡 Read the matching answer choice.

[1] #9 4.OA.B.4 Recognize that "even multiple of $3$" $=$ "multiple of $2$ AND multiple of $3$"

[2] #5 4.OA.B.4 If $n^2$ is a multiple of $6$, then $n$ itself must be a multiple of $6$. Reason

[3] #2 5.NBT.B.5 List the squares in order until one passes $2020$. With $n = 6k$, compute $36k^2

[4] #2 4.NBT.A.2 Compare each entry to $2020$. The first seven ($36, 144, 324, 576, 900, 1296, 17

[5] #3 4.NBT.A.2 $7$ matches choice (A).

Review

Reasonableness: Cross-check with a bound argument: $36k^2 < 2020 \iff k^2 < 56.\overline{1} \iff k \leq 7$ (since $7^2 = 49 < 56$ but $8^2 = 64 > 56$). So $k$ ranges over $1, 2, \ldots, 7$ — exactly $7$ values. ✓ The listed squares $36, 144, 324, 576, 900, 1296, 1764$ are all even (end in even digits), all multiples of $3$ (digit sums $9, 9, 9, 18, 9, 18, 18$), and all under $2020$. ✓

Alternative: Tool #16 (Complement / Change Focus): instead of building from $n = 6k$, list all perfect squares $1, 4, 9, 16, \ldots, 1936, 2025$ less than $2020$ ($n = 1, \ldots, 44$), then keep only those where $n$ is a multiple of $6$ ($n = 6, 12, 18, 24, 30, 36, 42$ — and $48 > 44$). That gives the same $7$ values.

CCSS standards used (min grade 5)

4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite (Combining "multiple of $2$" and "multiple of $3$" into "multiple of $6$", and using the fact that primes in $n^2$ already appear in $n$.)
5.NBT.B.5 Fluently multiply multi-digit whole numbers (Computing the candidate squares $36 \cdot k^2$ for $k = 1, \ldots, 8$ (up to $36 \cdot 64 = 2304$).)
4.NBT.A.2 Read and write multi-digit whole numbers and compare using symbols (Comparing each computed square to $2020$ and matching the final count $7$ to choice (A).)

⭐ This AMC 10 problem only needs Grade 5 multiplication and factor reasoning you already know — "even multiple of $3$" just means "multiple of $6$", so list the squares $(6 \cdot 1)^2, (6 \cdot 2)^2, \ldots$ until one passes $2020$. The first seven ($36, 144, 324, 576, 900, 1296, 1764$) fit, the eighth ($2304$) doesn't — answer $7$.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: A

Review

CCSS standards used (min grade 5)

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