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2023 AMC 12A Problems/Problem 3

Revision as of 02:17, 7 July 2024 by Robindabank (talk | contribs) (→‎Solution 5 (Under 10 seconds, ignore the first paragraph))
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 $5$?

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

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

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

~kyogrexu (minor edits by vadava_lx) ~ It was just a worse way of describing solution 5, hence removed by ~ Dextrik

Solution 5 (Under 10 seconds, ignore the first paragraph)

The original way of BlueShardow

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 $\left \lfloor{\frac{44}{5}}\right \rfloor = \boxed{\textbf{(A) 8}}$ solutions. PLEASE DO NOT do this problem this way, it takes way too much time.

~BlueShardow

The way of BlueShardow refined:

All it takes is to recall that 45 squared is 2025, and 45 is 5 x 9. So all the squares of 5 x 1, 5 x 2, 5 x 3 so on are divisible by 5. So the answer is$\left \lfloor{\frac{44}{5}}\right \rfloor = \boxed{\textbf{(A) 8}}$. It can be done even if one does not remember that 45 squared is 2025, all it takes is intuition. One can easily see mentally that 5 x 8 that is 40 squared is 1600, and then one has to do just one more computation and see that 5 x 9 that is 45 squared exceeds 2023, so the answer is $\left \lfloor{\frac{44}{5}}\right \rfloor = \boxed{\textbf{(A) 8}}$. BlueShardow's method is the best but he did not realize it.

~edit by RobinDaBank

Video Solution (easy to understand) by Power Solve

https://youtu.be/YXIH3UbLqK8?si=aIYHWEU82uUu21fQ&t=165

Video Solution by Math-X (First understand the problem!!!)

https://youtu.be/cMgngeSmFCY?si=E0a8wvcNRoeg2A3X&t=422

Video Solution by CosineMethod [🔥Fast and Easy🔥]

https://www.youtube.com/watch?v=wNH6O8D-7dY

Video Solution

https://youtu.be/w7RBPIatRNE

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


Video Solution (🚀 Just 2 min 🚀)

https://youtu.be/Z3fmCkuHG3c

~Education, the Study of Everything

Video Solution (easy to digest) by Power Solve

https://www.youtube.com/watch?v=8huvzWTtgaU

Video Solution (Easy to Understand) by DR.GOOGLE (YT: Pablo's Math)

https://youtu.be/BNhRdnOu-jI

See Also

2023 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 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 by
Problem 2 		Followed by
Problem 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

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