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2023 AMC 10A Problem 3

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2023 AMC 10A Problem 3AMC 10 Step by Step

Name: Date: September 13, 2026

Number Properties · difficulty 1 of 5 · about 2 min · position P1-10

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

A)
$8$
B)
$9$
C)
$10$
D)
$11$
E)
$12$

Answer

Key insight

A square is divisible by 5 exactly when its root is, so count multiples of 5 whose square stays below 2023, i.e. up to 40.

Solution

Because $5$ is prime, $5$ divides $n^2$ exactly when $5$ divides $n$ . So we need to count positive multiples of $5$ whose squares are below $2023$ .

Since $44^2 = 1936 < 2023$ and $45^2 = 2025 > 2023$ , the roots allowed are $n = 1, 2, \ldots, 44$ . The multiples of $5$ in this range are $5, 10, 15, \ldots, 40$ , which is $8$ numbers.

Their squares are $25, 100, 225, \ldots, 1600$ , all below $2023$ and all divisible by $5$ .

The answer is $\boxed{\textbf{(A)}\ 8}$ .

Why this works

Divisibility of a square by a prime reduces to divisibility of the root, which turns a question about squares into a question about multiples. The only care needed is at the boundary: $45$ is a multiple of $5$ but $45^2 = 2025$ just misses the cutoff, and the problem is designed so that this exact case decides between (A) and (B).

The trap

Including 45^2 = 2025, which exceeds 2023, and answering 9.

Common mistakes

Including 45^2 = 2025, which exceeds 2023, and answering 9.
Counting squares divisible by $25$ separately or thinking a square divisible by $5$ needs only its last digit checked.

Techniques

Bound the quantity above/below or estimate to pin it down · Set up the equation/formula and compute; no special trick needed