Difference between revisions of "2023 AMC 12A Problems/Problem 3"
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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>. | 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>. | ||
| − | ~ | + | ~sohan |
==Solution 4== | ==Solution 4== | ||
Revision as of 16:40, 11 November 2023
- The following problem is from both the 2023 AMC 10A #3 and 2023 AMC 12A #3, so both problems redirect to this page.
Contents
Problem
How many positive perfect squares less than 
are divisible by 
?
Solution 1
Note that 
but 
(which is over our limit of ). Therefore, the list is 
. There are 
elements, so the answer is 
.
~zhenghua ~walmartbrian (Minor edits for clarity by Technodoggo)
Solution 2 (slightly refined)
Since 
, there are 
perfect squares less than 2023.
~not_slay
Solution 3 (even better)
Since is prime, each solution must be divisible by 
. We take 
and see that there are positive perfect squares no greater than 
.
~sohan
Solution 4
We know the highest value would be at least 
but less than 
so we check 
, prime factorizing 45. We get 
. We square this and get 
. We know that 
, then we add 25 and get 
, which does not satisfy our requirement of having the square less than . The largest multiple of that satisfies this is and the smallest multiple of that works is so all multiples of from to satisfy the requirements. Now we divide each element of the set by and get 
so there are solutions.
~kyogrexu (minor edits by vadava_lx)
Solution 5
If we want to have the square be divisible by we must have it such that it is at least divisible by 
, since every prime in it's prime factorization must have an even power.
So, we must have 
, and we see the range of x is 
.
Therefore, there are solutions.
~ESAOPS
Video Solution by Math-X (First understand the problem!!!)
https://youtu.be/cMgngeSmFCY?si=E0a8wvcNRoeg2A3X&t=422
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution
~Education, the Study of Everything
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. 
