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]]}} | ||
| − | ==Problem== | + | ==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>? | ||
Revision as of 06:37, 26 August 2024
- 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 3
How many positive perfect squares less than 
are divisible by 
?
Solution 2 (slightly refined)
Since 
, there are 
perfect squares less than 2023.
~not_slay
Solution 3
Since is square-free, each solution must be divisible by 
. We take 
and see that there are 
positive perfect squares no greater than 
.
~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 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
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution
~Education, the Study of Everything
Video Solution by Power Solve
https://www.youtube.com/watch?v=8huvzWTtgaU
Video Solution by Pablo's Math
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. 
