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Difference between revisions of "2023 AMC 12A Problems/Problem 3"
 
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{{duplicate|[[2023 AMC 10A Problems/Problem 3|2023 AMC 10A #3]] and [[2023 AMC 12A Problems/Problem 3|2023 AMC 12A #3]]}}
 
{{duplicate|[[2023 AMC 10A Problems/Problem 3|2023 AMC 10A #3]] and [[2023 AMC 12A Problems/Problem 3|2023 AMC 12A #3]]}}
  
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==Problem==
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==Problem 3==
 
How many positive perfect squares less than <math>2023</math> are divisible by <math>5</math>?
 
How many positive perfect squares less than <math>2023</math> are divisible by <math>5</math>?
  
 
<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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==Solution 1 (slightly refined)==
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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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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> perfect squares less than 2023 that are divisible by 5.
  
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~zhenghua
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~not_slay (edited a teeny bit by mihikamishra)
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~walmartbrian
 
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(Minor edits for clarity by Technodoggo)
 
  
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==Solution 2 (slightly refined)==
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==Solution 2 ==
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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> perfect squares less than 2023.
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Since <math>5</math> is square-free, each solution must be divisible by <math>5^2=25</math>. We take <math>\left \lfloor{\frac{2023}{25}}\right \rfloor = 80</math> and see that there are <math>\boxed{\textbf{(A) 8}}</math> positive perfect squares no greater than <math>80</math>.
  
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~not_slay
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~jwseph
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==Solution 3 ==
 +
 
 +
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 figure 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  <math>\left \lfloor{\frac{44}{5}}\right \rfloor = \boxed{\textbf{(A) 8}}</math> solutions.
 +
 
 +
~BlueShardow
 +
 
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==Solution 4 ==
 +
 
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The way of BlueShardow refined:
 +
 
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All it takes is to recall that 45 squared is 2025, and 45 is 5 x 9. So all the squares of 5 x 1, 5 x 2, 5 x 3 so on are divisible by 5. So the answer is 8. It can be done even if one does not remember that 45 squared is 2025, all it takes is intuition. One can easily see mentally that 5 x 8 that is 40 squared is 1600, and then one has to do just one more computation and see that 5 x 9 that is 45 squared exceeds 2023, so the answer is 8.
 +
 
 +
~edit by RobinDaBank
 +
 
 +
Note that you can find the square of any number that ends in 5 by taking the number 5 more than it and the number 5 less than it, multiplying those together, and adding 25. For example, to calculate the square of 45, you do 40 x 50 = 2000, and 2000 + 25 = 2025.
 +
 
 +
~note by amadeus1011, edited by mihikamishra
  
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==Solution 3 (even better)==
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==Video Solution by Math-X ==
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Since <math>5</math> is prime, each solution must be divisible by <math>5^2=25</math>. We take <math>\left \lfloor{\frac{2023}{25}}\right \rfloor = 80</math> and see that there are <math>\boxed{\textbf{(A) 8}}</math> perfect squares no greater than <math>80</math>.
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https://youtu.be/GP-DYudh5qU?si=rwUloGNfN7tcoG-8&t=502
  
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~jwseph
 
  
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==Video Solution by Math-X (First understand the problem!!!)==
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==Video Solution by Power Solve==
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https://youtu.be/cMgngeSmFCY?si=E0a8wvcNRoeg2A3X&t=422
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https://youtu.be/YXIH3UbLqK8?si=aIYHWEU82uUu21fQ&t=165
 +
 
 +
== Video Solution by CosineMethod ==
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 +
https://www.youtube.com/watch?v=wNH6O8D-7dY
  
 
==Video Solution==
 
==Video Solution==
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
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 +
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==Video Solution ==
 +
https://youtu.be/Z3fmCkuHG3c
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 +
~Education, the Study of Everything
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 +
==Video Solution by Power Solve==
 +
https://www.youtube.com/watch?v=8huvzWTtgaU
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 +
==Video Solution by Pablo's Math==
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https://youtu.be/BNhRdnOu-jI
  
 
==See Also==
 
==See Also==

Latest revision as of 07:50, 6 September 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 3

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 (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 that are divisible by 5.

~not_slay (edited a teeny bit by mihikamishra)

Solution 2

Since $5$ is square-free, 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 3

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 figure 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 $\left \lfloor{\frac{44}{5}}\right \rfloor = \boxed{\textbf{(A) 8}}$ solutions.

~BlueShardow

Solution 4

The way of BlueShardow refined:

All it takes is to recall that 45 squared is 2025, and 45 is 5 x 9. So all the squares of 5 x 1, 5 x 2, 5 x 3 so on are divisible by 5. So the answer is 8. It can be done even if one does not remember that 45 squared is 2025, all it takes is intuition. One can easily see mentally that 5 x 8 that is 40 squared is 1600, and then one has to do just one more computation and see that 5 x 9 that is 45 squared exceeds 2023, so the answer is 8.

~edit by RobinDaBank

Note that you can find the square of any number that ends in 5 by taking the number 5 more than it and the number 5 less than it, multiplying those together, and adding 25. For example, to calculate the square of 45, you do 40 x 50 = 2000, and 2000 + 25 = 2025.

~note by amadeus1011, edited by mihikamishra

Video Solution by Math-X

https://youtu.be/GP-DYudh5qU?si=rwUloGNfN7tcoG-8&t=502


Video Solution by Power Solve

https://youtu.be/YXIH3UbLqK8?si=aIYHWEU82uUu21fQ&t=165

Video Solution by CosineMethod

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

https://youtu.be/Z3fmCkuHG3c

~Education, the Study of Everything

Video Solution by Power Solve

https://www.youtube.com/watch?v=8huvzWTtgaU

Video Solution by 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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