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Difference between revisions of "2023 AMC 12A Problems/Problem 3"

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==Video Solution by Math-X (First understand the problem!!!)==
 
==Video Solution by Math-X (First understand the problem!!!)==
 
https://youtu.be/cMgngeSmFCY?si=E0a8wvcNRoeg2A3X&t=422
 
https://youtu.be/cMgngeSmFCY?si=E0a8wvcNRoeg2A3X&t=422
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==Video Solution==
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https://youtu.be/w7RBPIatRNE
 +
 +
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
  
 
==See Also==
 
==See Also==

Revision as of 18:45, 10 November 2023

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 1

Note that $40^2=1600$ but $45^{2}=2025$ (which is over our limit of $2023$). Therefore, the list is $5^2,10^2,15^2,20^2,25^2,30^2,35^2,40^2$. There are $8$ elements, so the answer is $\boxed{\textbf{(A) 8}}$.

~zhenghua ~walmartbrian (Minor edits for clarity by Technodoggo)

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


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

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

Video Solution

https://youtu.be/w7RBPIatRNE

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

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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