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

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<math>\textbf{(A) } 8 \qquad\textbf{(B) }9 \qquad\textbf{(C) }10 \qquad\textbf{(D) }11 \qquad\textbf{(E) } 12</math>
 
<math>\textbf{(A) } 8 \qquad\textbf{(B) }9 \qquad\textbf{(C) }10 \qquad\textbf{(D) }11 \qquad\textbf{(E) } 12</math>
  
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==Solution 1==
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trash players
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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>). Therefore, 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>.
 
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~zhenghua
 
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~walmartbrian
 
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(Minor edits for clarity by Technodoggo)
 
  
 
==Solution 2 (slightly refined)==
 
==Solution 2 (slightly refined)==
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~kyogrexu (minor edits by vadava_lx)
 
~kyogrexu (minor edits by vadava_lx)
  
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==Solution 5==
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==Solution 5 (DO NOT DO ON AN ACTUAL TEST)==
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If we want to have the square be divisible by <math>5</math> we must have it such that it is at least divisible by <math>25</math>, since every prime in it's prime factorization must have an even power.
 
  
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So, we must have <math>0<25x^2<2023</math>, and we see the range of x is <math>1 \leq x \leq 8</math>.
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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 8 solutions. PLEASE DO NOT do this problem this way, it takes way too much time.
  
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Therefore, there are <math>\boxed{\textbf{(A) 8}}</math> solutions.
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~BlueShardow
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~ESAOPS
 
  
 
==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 by CosineMethod [🔥Fast and Easy🔥]==
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https://www.youtube.com/watch?v=wNH6O8D-7dY
  
 
==Video Solution==
 
==Video Solution==
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~Education, the Study of Everything
 
~Education, the Study of Everything
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==Video Solution (easy to digest) by Power Solve==
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https://www.youtube.com/watch?v=8huvzWTtgaU
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==Video Solution (Easy to Understand) by DR.GOOGLE (YT: Pablo's Math)==
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https://youtu.be/BNhRdnOu-jI
  
 
==See Also==
 
==See Also==

Latest revision as of 01:54, 12 May 2024

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 $5$?

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

trash players

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 (the best)

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

We know the highest value would be at least $40$ but less than $50$ so we check $45$, prime factorizing 45. We get $3^2 \cdot 5$. We square this and get $81 \cdot 25$. We know that $80 \cdot 25 = 2000$, then we add 25 and get $2025$, which does not satisfy our requirement of having the square less than $2023$. The largest multiple of $5$ that satisfies this is $40$ and the smallest multiple of $5$ that works is $5$ so all multiples of $5$ from $5$ to $40$ satisfy the requirements. Now we divide each element of the set by $5$ and get $1-8$ so there are $\boxed{\textbf{(A) 8}}$ solutions.

~kyogrexu (minor edits by vadava_lx)

Solution 5 (DO NOT DO ON AN ACTUAL TEST)

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 8 solutions. PLEASE DO NOT do this problem this way, it takes way too much time.

~BlueShardow

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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