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| − | | + | #redirect[[2023 AMC 12A Problems/Problem 3]] |
| − | ==Problem==
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| − | How many positive perfect squares less than <math>2023</math> are divisible by <math>5</math>?
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| − | <cmath>\textbf{(A) } 8 \qquad\textbf{(B) }9 \qquad\textbf{(C) }10 \qquad\textbf{(D) }11 \qquad\textbf{(E) } 12</cmath>
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| − | ==Solution 1==
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| − | 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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| − | ~zhenghua
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| − | ==Solution 2 (slightly refined)==
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| − | Since <math>\left \lfloor{\sqrt{2023}}\right \rfloor = 44</math>, there are <math>\left \lfloor{\frac{44}{5}}\right \rfloor = \boxed{\textbf{(A) 8}}</math> positive integers.
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| − | ~not_slay
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| − | ==See Also==
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| − | {{AMC12 box|year=2023|ab=A|num-b=2|num-a=4}}
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| − | {{AMC10 box|year=2023|ab=A|num-b=2|num-a=4}}
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| − | {{MAA Notice}}
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