2023 AMC 12A Problems/Problem 3
- The following problem is from both the 2023 AMC 10A #3 and 2023 AMC 12A #3, so both problems redirect to this page.
Contents
- 1 Problem 3
- 2 Solution 1 (slightly refined)
- 3 Solution 2
- 4 Solution 3
- 5 Solution 4
- 6 Video Solution by Little Fermat
- 7 Video Solution by Math-X
- 8 Video Solution by Power Solve
- 9 Video Solution by CosineMethod
- 10 Video Solution
- 11 Video Solution
- 12 Video Solution by Power Solve
- 13 Video Solution by Pablo's Math
- 14 See Also
Problem 3
How many positive perfect squares less than 
are divisible by 
?
Solution 1 (slightly refined)
Since 
, there are 
perfect squares less than 2023 that are divisible by 5.
Solution 2
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 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 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 
to 
, and the multiples of five are {, 
..., 
, 
}, dividing each of them by , you get {, 
..., 
, 
}, therefore, the answer is
~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
~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. 
