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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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==Solution 2 ==
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~walmartbrian
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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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(Minor edits for clarity by Technodoggo)
 
  
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==Solution 2 (slightly refined)==
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~jwseph
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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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==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.  
 +
 
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~BlueShardow
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==Solution 4 ==
 +
 
 +
The way of BlueShardow refined:
  
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~not_slay
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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.
  
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==Solution 3 (even better)==
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~edit by RobinDaBank
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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> positive perfect squares no greater than <math>80</math>.
 
  
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~jwseph
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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.
  
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==Solution 4==
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And then you know that it goes up from <math>1</math> to <math>44</math>, and the multiples of five are {<math>5</math>, <math>10</math>..., <math>35</math>, <math>40</math>}, dividing each of them by <math>5</math>, you get {<math>1</math>, <math>2</math>..., <math>7</math>, <math>8</math>}, therefore, the answer is <math>\boxed{\textbf{(A) 8}}</math>
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We know the highest value would be at least <math>40</math> but less than <math>50</math> so we check <math>45</math>, prime factorizing 45. We get <math>3^2 \cdot 5</math>. We square this and get <math>81 \cdot 25</math>. We know that <math>80 \cdot 25 = 2000</math>, then we add 25 and get <math>2025</math>, which does not satisfy our requirement of having the square less than <math>2023</math>. The largest multiple of <math>5</math> that satisfies this is <math>40</math> and the smallest multiple of <math>5</math> that works is <math>5</math> so all multiples of <math>5</math> from <math>5</math> to <math>40</math> satisfy the requirements. Now we divide each element of the set by <math>5</math> and get <math>1-8</math> so there are <math>\boxed{\textbf{(A) 8}}</math> solutions.
 
  
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~kyogrexu (minor edits by vadava_lx)
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~note by amadeus1011, edited by mihikamishra, then edited by ~NXC
  
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==Solution 5==
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==Video Solution by Little Fermat==
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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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https://youtu.be/h2Pf2hvF1wE?si=3FSk6sTPCUg9J0U1&t=394
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~little-fermat
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==Video Solution by Math-X ==
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https://youtu.be/GP-DYudh5qU?si=rwUloGNfN7tcoG-8&t=502
  
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So, we must have <math>0<25x^2<2023</math>, and we see the range of x is <math>1\seq x^2\leq 8</math>.
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==Video Solution by Power Solve==
 +
https://youtu.be/YXIH3UbLqK8?si=aIYHWEU82uUu21fQ&t=165
  
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Therefore, there are <math>\boxed{\textbf{(A) 8}}</math> solutions.
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== Video Solution by CosineMethod ==
  
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==Video Solution by Math-X (First understand the problem!!!)==
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https://www.youtube.com/watch?v=wNH6O8D-7dY
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https://youtu.be/cMgngeSmFCY?si=E0a8wvcNRoeg2A3X&t=422
 
  
 
==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 ==
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https://youtu.be/Z3fmCkuHG3c
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~Education, the Study of Everything
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 +
==Video Solution by Power Solve==
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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 06:17, 5 November 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.

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.

And then you know that it goes up from $1$ to $44$, and the multiples of five are {$5$, $10$..., $35$, $40$}, dividing each of them by $5$, you get {$1$, $2$..., $7$, $8$}, therefore, the answer is $\boxed{\textbf{(A) 8}}$

~note by amadeus1011, edited by mihikamishra, then edited by ~NXC

Video Solution by Little Fermat

https://youtu.be/h2Pf2hvF1wE?si=3FSk6sTPCUg9J0U1&t=394 ~little-fermat

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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