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)
m (→‎Solution 5)
(27 intermediate revisions by 13 users not shown)
Line 5: Line 5:
  
 
<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>
−
 
−
==Solution 1==
 
−
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>.
 
−
 
−
~zhenghua
 
−
~walmartbrian
 
−
(Minor edits for clarity by Technodoggo)
 
  
 
==Solution 2 (slightly refined)==
 
==Solution 2 (slightly refined)==
Line 18: Line 11:
 
~not_slay
 
~not_slay
  
−
==Solution 3 (even better)==
+
==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 ==
−
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.
+
 
 +
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.  
  
−
~kyogrexu (minor edits by vadava_lx)
+
~BlueShardow
  
−
==Solution 5==
+
==Solution 5 ==
−
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.
 
  
−
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>.
+
The way of BlueShardow refined:
  
−
Therefore, there are <math>\boxed{\textbf{(A) 8}}</math> solutions.
+
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.  
  
−
~ESAOPS
+
~edit by RobinDaBank
  
−
==Video Solution by Math-X (First understand the problem!!!)==
+
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
 
https://youtu.be/cMgngeSmFCY?si=E0a8wvcNRoeg2A3X&t=422
 +
 +
== Video Solution by CosineMethod ==
 +
 +
https://www.youtube.com/watch?v=wNH6O8D-7dY
  
 
==Video Solution==
 
==Video Solution==
Line 45: Line 49:
  
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 
~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==
 
==See Also==

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"