Difference between revisions of "2023 AMC 10A 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>\text{\boxed{(A)}}</math>
 
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>\text{\boxed{(A)}}</math>
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==See Also==
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{{AMC10 box|year=2023|ab=A|before=2|num-a=4}}

Revision as of 14:59, 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 $45^{2}=2025$ so the list is $5,10,15,20,25,30,35,40$ there are $8$ elements so the answer is $\text{\boxed{(A)}}$

See Also

2023 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
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