# 2023 AMC 12A Problems/Problem 3

The following problem is from both the 2023 AMC 10A #3 and 2023 AMC 12A #3, so both problems redirect to this page.

## Problem

How many positive perfect squares less than $2023$ are divisible by $2$?

$\textbf{(A) } 8 \qquad\textbf{(B) }9 \qquad\textbf{(C) }10 \qquad\textbf{(D) }11 \qquad\textbf{(E) } 12$

trash players

## Solution 2 (slightly refined)

Since $\left \lfloor{\sqrt{2023}}\right \rfloor = 44$, there are $\left \lfloor{\frac{44}{5}}\right \rfloor = \boxed{\textbf{(A) 8}}$ perfect squares less than 2023.

~not_slay

## Solution 3 (the best)

Since $5$ is prime, each solution must be divisible by $5^2=25$. We take $\left \lfloor{\frac{2023}{25}}\right \rfloor = 80$ and see that there are $\boxed{\textbf{(A) 8}}$ positive perfect squares no greater than $80$.

~jwseph

## Solution 4

We know the highest value would be at least $40$ but less than $50$ so we check $45$, prime factorizing 45. We get $3^2 \cdot 5$. We square this and get $81 \cdot 25$. We know that $80 \cdot 25 = 2000$, then we add 25 and get $2025$, which does not satisfy our requirement of having the square less than $2023$. The largest multiple of $5$ that satisfies this is $40$ and the smallest multiple of $5$ that works is $5$ so all multiples of $5$ from $5$ to $40$ satisfy the requirements. Now we divide each element of the set by $5$ and get $1-8$ so there are $\boxed{\textbf{(A) 8}}$ solutions.

~kyogrexu (minor edits by vadava_lx)

## Solution 5 (DO NOT DO ON AN ACTUAL TEST)

Since the perfect squares have to be divisible by 5, then we know it has to be 5 times some number squared (5*x)^2. With this information, you can fique out every single product of 5 and another number squared to count how many perfect squares are divisible by 5 that are less than 2023. (EX: 5^2 = 25, 10^2 = 100, ... 40^2 = 1600) With this you get a max of 40^2, or 8 solutions. PLEASE DO NOT do this problem this way, it takes way too much time.

~BlueShardow

## Video Solution

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

## Video Solution (🚀 Just 2 min 🚀)

~Education, the Study of Everything

## See Also

 2023 AMC 10A (Problems • Answer Key • Resources) Preceded byProblem 2 Followed byProblem 4 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions
 2023 AMC 12A (Problems • Answer Key • Resources) Preceded byProblem 2 Followed byProblem 4 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.