AMC 10 · 2020 · #7
Easy mode Grade 5Problem
We are looking at positive whole numbers smaller than 2020.
We want the numbers that fit all three of these rules:
- the number is a multiple of 3,
- the number is even,
- the number is a perfect square (like 1,4,9,16,25,…).
How many such numbers are there?
Pick an answer.
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.
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 $3$.
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) $12$
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
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 $6$.
💡 Simpler phrasing: just multiples of $6$.
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 $n^2 < 2020$.
💡 $n^2$ multiple of $6$ $\Rightarrow$ $n$ multiple of $6$.
5.NBT.B.5 Step 3 - List the squares in order until one passes $2020$.
- With $n = 6k$, compute $36k^2$ for $k = 1, 2, 3, \ldots$
💡 Generate candidates in order from smallest.
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.
💡 Stop the list at the first miss.
4.NBT.A.2 Step 5 $7$ matches choice (A).
💡 Read the matching answer choice.
4.OA.B.4 Recognize that "even multiple of $3$" $=$ "multiple of $2$ AND multiple of $3$" 4.OA.B.4 If $n^2$ is a multiple of $6$, then $n$ itself must be a multiple of $6$. Reason 5.NBT.B.5 List the squares in order until one passes $2020$. With $n = 6k$, compute $36k^2 4.NBT.A.2 Compare each entry to $2020$. The first seven ($36, 144, 324, 576, 900, 1296, 17 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.4Find 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.5Fluently 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.2Read 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 $7$.
⭐ 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 $7$.
More like this
Same archetype — closest grade level first.