Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 5 (Under 10 seconds, ignore the first paragraph))
m (→‎Solution 5)
(3 intermediate revisions by 2 users not shown)
Line 12: Line 12:
  
 
==Solution 3 ==
 
==Solution 3 ==
−
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>.
+
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>.
  
 
~jwseph
 
~jwseph
  
−
==Solution 4==
+
==Solution 4 ==
  
−
~kyogrexu (minor edits by vadava_lx)
+
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.
−
~ It was just a worse way of describing solution 5, hence removed by
 
−
~ Dextrik
 
  
−
==Solution 5 (Under 10 seconds, ignore the first paragraph)==
+
~BlueShardow
  
−
The original way of BlueShardow
+
==Solution 5 ==
  
−
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  <math>\left \lfloor{\frac{44}{5}}\right \rfloor = \boxed{\textbf{(A) 8}}</math> solutions. PLEASE DO NOT do this problem this way, it takes way too much time.
+
The way of BlueShardow refined:
  
−
~BlueShardow
+
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.
  
−
The way of BlueShardow refined:
+
~edit by RobinDaBank
  
−
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. BlueShardow's method is the best but he did not realize it.
+
Note that you can find the square of any number that ends in 5 by taking all the numbers but the last one (let's call them A), then multiplying A(A+1). In the end, addend 25 to the end of the number. For example, 4 x 5 = 20. Therefore, 45 squared is 20 25.
  
−
~edit by RobinDaBank
+
~note by amadeus1011
  
−
==Video Solution (easy to understand) by Power Solve==
+
==Video Solution by Power Solve==
 
https://youtu.be/YXIH3UbLqK8?si=aIYHWEU82uUu21fQ&t=165
 
https://youtu.be/YXIH3UbLqK8?si=aIYHWEU82uUu21fQ&t=165
  
−
==Video Solution by Math-X (First understand the problem!!!)==
+
==Video Solution by Math-X ==
 
https://youtu.be/cMgngeSmFCY?si=E0a8wvcNRoeg2A3X&t=422
 
https://youtu.be/cMgngeSmFCY?si=E0a8wvcNRoeg2A3X&t=422
  
−
== Video Solution by CosineMethod [🔥Fast and Easy🔥]==
+
== Video Solution by CosineMethod ==
  
 
https://www.youtube.com/watch?v=wNH6O8D-7dY
 
https://www.youtube.com/watch?v=wNH6O8D-7dY
Line 53: Line 51:
  
  
−
==Video Solution (🚀 Just 2 min 🚀)==
+
==Video Solution ==
 
https://youtu.be/Z3fmCkuHG3c
 
https://youtu.be/Z3fmCkuHG3c
  
 
~Education, the Study of Everything
 
~Education, the Study of Everything
  
−
==Video Solution (easy to digest) by Power Solve==
+
==Video Solution by Power Solve==
 
https://www.youtube.com/watch?v=8huvzWTtgaU
 
https://www.youtube.com/watch?v=8huvzWTtgaU
  
−
==Video Solution (Easy to Understand) by DR.GOOGLE (YT: Pablo's Math)==
+
==Video Solution by Pablo's Math==
 
https://youtu.be/BNhRdnOu-jI
 
https://youtu.be/BNhRdnOu-jI
  

Revision as of 15:52, 22 July 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 $5$?

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

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

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 4

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 5

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 all the numbers but the last one (let's call them A), then multiplying A(A+1). In the end, addend 25 to the end of the number. For example, 4 x 5 = 20. Therefore, 45 squared is 20 25.

~note by amadeus1011

Video Solution by Power Solve

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

Video Solution by Math-X

https://youtu.be/cMgngeSmFCY?si=E0a8wvcNRoeg2A3X&t=422

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2023_AMC_12A_Problems/Problem_3&oldid=223406"