AMC 10 · 2023 · #3

Grade 6 number-theory

perfect-squaresdivisibility-rulesprime-numbersmultiples pattern-recognitionbound-inequality-then-enumerate ↑ Prerequisites: perfect-squaresmultiples

📏 Short solution 💡 2 insights

Problem

How many positive perfect squares less than $2023$ are divisible by $5$ ?

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: Count the positive perfect squares that are (a) less than $2023$ and (b) divisible by $5$.

Givens: We want $N = k^2$ for some positive integer $k$; Condition 1: $N < 2023$; Condition 2: $5 \mid N$; Answer choices: (A) $8$, (B) $9$, (C) $10$, (D) $11$, (E) $12$

Unknowns: The number of positive perfect squares $N$ satisfying both conditions

Understand

Restated: Count the positive perfect squares that are (a) less than $2023$ and (b) divisible by $5$.

Givens: We want $N = k^2$ for some positive integer $k$; Condition 1: $N < 2023$; Condition 2: $5 \mid N$; Answer choices: (A) $8$, (B) $9$, (C) $10$, (D) $11$, (E) $12$

Plan

Primary tool: #5 Look for a Pattern

Secondary: #2 Make a Systematic List

After noticing that $5 \mid k^2$ forces $5 \mid k$ (because $5$ is prime), the qualifying squares are exactly $5^2, 10^2, 15^2, \ldots$ — a perfectly regular arithmetic-style pattern in $k$. Tool #5 (Look for a Pattern) frames it: write $k = 5m$ and the constraint becomes $25 m^2 < 2023$. Tool #2 (Systematic List) then finishes by listing $m = 1, 2, 3, \ldots$ in order and stopping at the first $m$ that breaks the bound. Algebra (#13) would be heavier and a brute list of all $44$ squares under $2023$ is wasteful; the pattern + list combo is the natural fit.

Execute — Answer: A

#5 Look for a Pattern 6.EE.A.1 Step 1

Use prime factorization to tighten the divisibility condition.
A perfect square's prime factorization has even exponents, so if $5$ divides $k^2$ it must appear with exponent at least $2$, which means $5$ already divides $k$.
So $k$ must be a multiple of $5$.

$$5 \mid k^2 \;\Longleftrightarrow\; 5 \mid k$$

💡 Reading a square through its prime factors shows that any prime dividing the square already had to divide the base — the Grade 6 "whole-number exponent" idea.

#5 Look for a Pattern 6.EE.B.6 Step 2

Write $k = 5m$ for a positive integer $m$, so that $N = k^2 = 25 m^2$.
The constraint $N < 2023$ becomes a constraint on $m^2$ alone.

$$N = (5m)^2 = 25 m^2 < 2023 \;\Longrightarrow\; m^2 < \dfrac{2023}{25} = 80.92$$

💡 Renaming $k$ as $5m$ turns the pattern "$k$ is a multiple of $5$" into a plain Grade 6 expression in $m$, with no divisibility left to track.

#2 Make a Systematic List 3.OA.C.7 Step 3

List the positive integers $m$ in order with their squares and stop when $m^2 > 80.92$.
This is fast because the squares are familiar.

$$1^2{=}1,\;2^2{=}4,\;3^2{=}9,\;4^2{=}16,\;5^2{=}25,\;6^2{=}36,\;7^2{=}49,\;8^2{=}64,\;9^2{=}81$$

💡 Knowing $1^2$ through $9^2$ by heart is the Grade 3 multiplication-fluency standard — the list assembles itself.

#2 Make a Systematic List 4.OA.B.4 Step 4

Compare to the bound $80.92$.
The values $1$ through $64$ are all under $80.92$, so $m = 1, 2, \ldots, 8$ all work.
But $9^2 = 81 > 80.92$, so $m = 9$ fails.
The valid $m$ form the set $\{1, 2, 3, 4, 5, 6, 7, 8\}$ — exactly $8$ values.

$$m \in \{1, 2, 3, 4, 5, 6, 7, 8\} \;\Rightarrow\; 8 \text{ values} \;\Rightarrow\; \textbf{(A)}$$

💡 Each $m$ gives one allowed $k = 5m$ and hence one perfect square $25 m^2$, so counting valid $m$ is the same as counting valid $N$ — a Grade 4 factor-pair / multiples-style count.

[1] #5 6.EE.A.1 Use prime factorization to tighten the divisibility condition. A perfect square'

[2] #5 6.EE.B.6 Write $k = 5m$ for a positive integer $m$, so that $N = k^2 = 25 m^2$. The const

[3] #2 3.OA.C.7 List the positive integers $m$ in order with their squares and stop when $m^2 >

[4] #2 4.OA.B.4 Compare to the bound $80.92$. The values $1$ through $64$ are all under $80.92$,

Review

Reasonableness: Spot-check the endpoints. $m = 1$ gives $N = 25$, which is $5^2$, less than $2023$, and divisible by $5$ — qualifies. $m = 8$ gives $N = 25 \cdot 64 = 1600 = 40^2$, less than $2023$, divisible by $5$ — qualifies. $m = 9$ gives $N = 25 \cdot 81 = 2025 \geq 2023$ — fails, confirming the cap. So the answer is exactly $8$, matching choice (A).

Alternative: Tool #6 (Guess and Check) on the upper bound. We need $k^2 < 2023$ with $5 \mid k$, so test the largest multiple of $5$ whose square stays under $2023$. $\sqrt{2023} \approx 44.98$, so $k \leq 44$ and the largest multiple of $5$ within that is $k = 40$. Listing the multiples $5, 10, 15, 20, 25, 30, 35, 40$ gives $8$ values, matching (A).

CCSS standards used (min grade 6)

3.OA.C.7 Fluently multiply and divide within 100 (Knowing $1^2$ through $9^2$ at sight so the list $1, 4, 9, \ldots, 81$ can be written down in one pass and compared to $80.92$.)
4.OA.B.4 Find all factor pairs for a whole number; recognize multiples; determine prime or composite (Counting the qualifying values of $m$ as a multiples-of-$5$ count for $k$, i.e. $k \in \{5, 10, 15, \ldots, 40\}$.)
6.EE.A.1 Write and evaluate numerical expressions involving whole-number exponents (Applying "prime in a square forces the prime in the base" — i.e. reasoning about exponents in $k^2$ to conclude $5 \mid k$.)
6.EE.B.6 Use variables to represent numbers and write expressions to solve problems (Letting $k = 5m$ and rewriting the constraint as $25 m^2 < 2023$, i.e. $m^2 < 80.92$.)

⭐ This AMC 10 problem only needs Grade 6 "$5$ inside a square means $5$ inside its root" reasoning — the squares of $5, 10, \ldots, 40$ are exactly the eight that fit.