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

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==Solution 1==
 
==Solution 1==
Note that <math>45^{2}=2025</math> so the list is <math>5,10,15,20,25,30,35,40</math> there are <math>8</math> elements so the answer is <math>\boxed{\textbf{(A) 8}}</math>.
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Note that <math>40^2=1600</math> but <math>45^{2}=2025</math> (which is over our limit of <math>2023</math>) so the list is <math>5^2,10^2,15^2,20^2,25^2,30^2,35^2,40^2</math>. There are <math>8</math> elements, so the answer is <math>\boxed{\textbf{(A) 8}}</math>.  
  
 
~zhenghua
 
~zhenghua
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 +
(Minor edits for clarity by Technodoggo)
  
 
==Solution 2 (slightly refined)==
 
==Solution 2 (slightly refined)==

Revision as of 22:50, 9 November 2023

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 $2023$) so 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

(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

See Also

2023 AMC 10A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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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